Mechanics Alıştırma Soruları — JAMB UTME — Sayfa 8

Soru 141Soru →

A body of mass $2.5\text{ kg}$ moving at a speed of $4.0\text{ m s}^{-1}$ along a straight horizontal path is brought to rest in $2.0\text{ s}$ by a constant retarding force. What is the magnitude of this retarding force in newtons?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The magnitude of the retarding force is

5.0\text{ N}

.

According to Newton's second law of motion, net force is equal to the rate of change of momentum (

F = \frac{\Delta p}{\Delta t} = \frac{m(v-u)}{t}

). Substituting

m = 2.5\text{ kg}

,

u = 4.0\text{ m s}^{-1}

,

v = 0\text{ m s}^{-1}

, and

t = 2.0\text{ s}

gives

F = \frac{2.5 \times (0 - 4.0)}{2.0} = -5.0\text{ N}

. The magnitude of this force is

5.0\text{ N}

.

Adım Adım Çözüm

1

Determine the initial momentum and final momentum of the body.

Initial momentum

p_i = 2.5 \times 4.0 = 10.0\text{ kg m s}^{-1}

, and final momentum

p_f = 0\text{ kg m s}^{-1}

.

Linear momentum is defined as the product of mass and velocity (

p = mv

).

2

Calculate the magnitude of the force applied using the impulse-momentum relationship

F = \frac{\Delta p}{\Delta t}

.

Magnitude of force

F = \frac{|0 - 10.0|}{2.0} = 5.0\text{ N}

.

Newton's second law states that the rate of change of momentum is equal to the net external force applied.

Anahtar Kavram

Newton's Second Law and Impulse-Momentum Relationship

Tahmini Süre:45s

Soru 142Soru →

A constant horizontal force of $25\text{ N}$ is applied to push a box across a smooth horizontal surface through a displacement of $8\text{ m}$ in the direction of the force. What is the total work done on the box?

200\text{ J}

40\text{ J}

3.125\text{ J}

33\text{ J}

Cevabı ve açıklamayı göster

Cevap: $200\text{ J}$

Cevap

The work done on the box is

200\text{ J}

.

Work done is defined as the product of force and distance moved in the direction of the force (

W = F \times d

). Substituting the given values

F = 25\text{ N}

and

d = 8\text{ m}

gives

W = 200\text{ J}

.

Adım Adım Çözüm

1

Identify the given physical quantities

Force

F = 25\text{ N}

, displacement

d = 8\text{ m}

, angle

\theta = 0^\circ

Work depends on force and displacement along the line of action

2

Apply the work formula

W = F \cdot d \cos(\theta)

W = 25\text{ N} \times 8\text{ m} \times \cos(0^\circ) = 200\text{ J}

Since force and displacement are in the same direction,

\cos(0^\circ) = 1

Anahtar Kavram

Work Done by a Constant Force

Soru 143Soru →

A body of mass $3.0\text{ kg}$ moving due east along a smooth horizontal track at $8.0\text{ m s}^{-1}$ collides head-on with a $2.0\text{ kg}$ body moving due west at $4.0\text{ m s}^{-1}$ . Immediately after the impact, the $2.0\text{ kg}$ body rebounds due east with a speed of $5.0\text{ m s}^{-1}$ . If the duration of the impact is $0.02\text{ s}$ , what is the magnitude of the average impact force exerted on the $3.0\text{ kg}$ body?

100\text{ N}

1200\text{ N}

900\text{ N}

1500\text{ N}

Cevabı ve açıklamayı göster

Cevap: $900\text{ N}$

Cevap

The magnitude of the average impact force exerted on the

3.0\text{ kg}

body is

900\text{ N}

.

By designating East as positive, initial velocities are

u_1 = +8.0\text{ m s}^{-1}

and

u_2 = -4.0\text{ m s}^{-1}

. Conservation of linear momentum gives

(3.0 \times 8.0) + (2.0 \times -4.0) = 3.0 v_1 + (2.0 \times 5.0)

, yielding

v_1 = +2.0\text{ m s}^{-1}

. The change in momentum of the

3.0\text{ kg}

body is

\Delta p = 3.0 \times (2.0 - 8.0) = -18.0\text{ N s}

. Dividing the magnitude of this impulse by the impact time (

0.02\text{ s}

) yields an average force of

900\text{ N}

.

Adım Adım Çözüm

1

Set up momentum conservation by assigning directional signs to velocities.

Taking East as positive (

+

):

u_1 = +8.0\text{ m s}^{-1}

,

u_2 = -4.0\text{ m s}^{-1}

,

v_2 = +5.0\text{ m s}^{-1}

.

Linear momentum is a vector quantity, so direction must be taken into account.

2

Calculate the initial total momentum and solve for the final velocity of the

3.0\text{ kg}

body (

v_1

).

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies (3.0 \times 8.0) + (2.0 \times -4.0) = 3.0 v_1 + (2.0 \times 5.0) \implies 16.0 = 3.0 v_1 + 10.0 \implies v_1 = +2.0\text{ m s}^{-1}

.

Total momentum is conserved in the absence of external forces.

3

Calculate the magnitude of the impulse and average force acting on the

3.0\text{ kg}

body.

\Delta p_1 = m_1(v_1 - u_1) = 3.0 \times (2.0 - 8.0) = -18.0\text{ N s}

. Force magnitude

F = \frac{|\Delta p_1|}{\Delta t} = \frac{18.0}{0.02} = 900\text{ N}

.

The average force equals the rate of change of linear momentum.

Anahtar Kavram

Law of Conservation of Linear Momentum and Impulse-Momentum Theorem

Tahmini Süre:2m 0s

Soru 144Soru →

An electric train traveling along a straight track uniformly slows down from a speed of $30\text{ m/s}$ to $10\text{ m/s}$ over a distance of $200\text{ m}$ . What is the magnitude of the deceleration of the train in $\text{m/s}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The magnitude of the deceleration is

2\text{ m/s}^2

.

Using the kinematic equation

v^2 = u^2 + 2as

, substituting

v = 10\text{ m/s}

,

u = 30\text{ m/s}

, and

s = 200\text{ m}

gives

100 = 900 + 400a

, which simplifies to

400a = -800

, yielding

a = -2\text{ m/s}^2

. The magnitude of deceleration is

2\text{ m/s}^2

.

Adım Adım Çözüm

1

Identify the given kinematic variables.

Initial velocity

u = 30\text{ m/s}

, final velocity

v = 10\text{ m/s}

, displacement

s = 200\text{ m}

.

Choosing the appropriate equation of motion requires knowing which variables are given and which is unknown.

2

Apply the third equation of motion relating initial velocity, final velocity, acceleration, and distance.

v^2 = u^2 + 2as

This formula connects

u

,

v

,

a

, and

s

without needing time

t

.

3

Substitute the given values into the equation and solve for acceleration

a

.

(10)^2 = (30)^2 + 2(a)(200) \implies 100 = 900 + 400a \implies 400a = -800 \implies a = -2\text{ m/s}^2

.

Performing algebraic operations to isolate the acceleration parameter.

4

State the magnitude of the deceleration.

The magnitude of deceleration is

2\text{ m/s}^2

.

Deceleration represents the rate of speed reduction, which corresponds to the magnitude of negative acceleration.

Anahtar Kavram

Uniformly Accelerated Motion Equations

Soru 145Soru →

Car $A$ , traveling at a constant speed of $20\text{ m/s}$ along a straight horizontal road, passes a landmark $50\text{ m}$ ahead of car $B$ , which is initially stationary. At $t = 0\text{ s}$ , car $B$ starts moving in the same direction, accelerating uniformly at $3.0\text{ m/s}^2$ until it reaches a top speed of $30\text{ m/s}$ , after which it continues at this constant top speed. How many seconds after $t = 0\text{ s}$ does car $B$ catch up with car $A$ ?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

Car B catches up with car A after 20 seconds.

Car B accelerates for

10\text{ s}

covering

150\text{ m}

to reach

30\text{ m/s}

. During these

10\text{ s}

, car A reaches a position of

250\text{ m}

(taking into account its

50\text{ m}

head start). Car B then closes the remaining

100\text{ m}

gap at a relative speed of

10\text{ m/s}

(

30\text{ m/s} - 20\text{ m/s}

), taking an extra

10\text{ s}

. The total elapsed time is

10\text{ s} + 10\text{ s} = 20\text{ s}

.

Adım Adım Çözüm

1

Calculate the duration

t_1

of car B's acceleration phase to reach

30\text{ m/s}

.

t_1 = \frac{v_{max} - u}{a} = \frac{30 - 0}{3.0} = 10\text{ s}

Car B accelerates uniformly from rest at

3.0\text{ m/s}^2

until reaching its top speed limit.

2

Determine the distance

s_B

covered by car B and the position

s_A

of car A at

t = 10\text{ s}

.

s_B = \frac{1}{2}a t_1^2 = \frac{1}{2}(3.0)(10)^2 = 150\text{ m}

;

s_A = 50 + v_A t_1 = 50 + (20)(10) = 250\text{ m}

Car A starts

50\text{ m}

ahead and moves continuously at

20\text{ m/s}

.

3

Find the separation distance between the two cars at

t = 10\text{ s}

and compute the time

\Delta t

required to close it.

\text{Separation} = 250 - 150 = 100\text{ m}; \Delta t = \frac{100}{30 - 20} = 10\text{ s}

Beyond

t = 10\text{ s}

, car B travels at a constant relative velocity of

10\text{ m/s}

faster than car A.

4

Sum the acceleration time and constant speed time to find the total time taken.

t_{total} = t_1 + \Delta t = 10\text{ s} + 10\text{ s} = 20\text{ s}

Combining both phases yields the exact instant car B overtakes car A.

Anahtar Kavram

Multi-stage relative motion with acceleration limits

Soru 146Soru →

A solid block floats in water of density $1000\text{ kg/m}^3$ with $60\%$ of its total volume submerged. When the same block is placed in an unknown liquid $X$ , $80\%$ of its total volume is submerged. What is the density of liquid $X$ ?

750\text{ kg/m}^3

1333\text{ kg/m}^3

480\text{ kg/m}^3

800\text{ kg/m}^3

Cevabı ve açıklamayı göster

Cevap: $750\text{ kg/m}^3$

Cevap

The density of liquid

X

is

750\text{ kg/m}^3

.

For any floating body, its weight equals the upthrust exerted by the liquid. The upthrust is given by

U = \rho_{\text{fluid}} \cdot V_{\text{submerged}} \cdot g

. Since the weight of the block is unchanged,

\rho_{\text{water}} \cdot V_{\text{sub, water}} = \rho_X \cdot V_{\text{sub, X}}

. Substituting the given values:

1000 \times 0.60V = \rho_X \times 0.80V

, yielding

\rho_X = 750\text{ kg/m}^3

.

Adım Adım Çözüm

1

Apply the Law of Flotation to the block in water

Weight of block

W = \rho_w \cdot V_{\text{sub, water}} \cdot g = 1000 \cdot 0.60V \cdot g = 600 V g

A floating object displaces its own weight of fluid.

2

Apply the Law of Flotation to the block in liquid X

Weight of block

W = \rho_X \cdot V_{\text{sub, X}} \cdot g = \rho_X \cdot 0.80V \cdot g

The weight of the block remains constant regardless of the fluid.

3

Equate the two expressions for the weight of the block and solve for

\rho_X

\rho_X \cdot 0.80V \cdot g = 600 V g \implies \rho_X = \frac{600}{0.80} = 750\text{ kg/m}^3

Since both buoyant forces equal the block's weight, set them equal to each other.

Anahtar Kavram

Law of Flotation and Archimedes' Principle

Soru 147Soru →

A solid object has a mass of $0.50\text{ kg}$ and a volume of $2.0 \times 10^{-4}\text{ m}^3$ . If it is completely immersed in a liquid of density $800\text{ kg/m}^3$ , what is the magnitude of the upthrust exerted on the object? [Take $g = 10\text{ m/s}^2$ ]

1.6\text{ N}

5.0\text{ N}

3.4\text{ N}

0.16\text{ N}

Cevabı ve açıklamayı göster

Cevap: $1.6\text{ N}$

Cevap

The magnitude of the upthrust exerted on the object is

1.6\text{ N}

.

By Archimedes' Principle, upthrust is equal to the weight of the liquid displaced:

U = \rho_{\text{liquid}} V g

. Substituting

\rho = 800\text{ kg/m}^3

,

V = 2.0 \times 10^{-4}\text{ m}^3

, and

g = 10\text{ m/s}^2

yields

1.6\text{ N}

.

Adım Adım Çözüm

1

Identify the given values and state Archimedes' Principle

Volume of displaced liquid

V = 2.0 \times 10^{-4}\text{ m}^3

, density of liquid

\rho = 800\text{ kg/m}^3

, acceleration due to gravity

g = 10\text{ m/s}^2

. Upthrust formula is

U = \rho V g

.

According to Archimedes' Principle, the upward buoyant force (upthrust) equals the weight of the liquid displaced by the submerged object.

2

Calculate the upthrust

U = 800\text{ kg/m}^3 \times (2.0 \times 10^{-4}\text{ m}^3) \times 10\text{ m/s}^2 = 1.6\text{ N}

.

Multiplying fluid density by submerged volume and gravitational acceleration gives the force in newtons.

Anahtar Kavram

Archimedes' Principle and Upthrust

Soru 148Soru →

A spherical ball bearing of radius $3.0\text{ mm}$ and density $5400\text{ kg/m}^3$ falls vertically through a viscous oil of density $900\text{ kg/m}^3$ and dynamic viscosity coefficient $0.10\text{ Pa}\cdot\text{s}$ . Assuming the motion obeys Stokes' law and taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the terminal velocity of the sphere?

0.90\text{ m/s}

1.08\text{ m/s}

0.68\text{ m/s}

0.18\text{ m/s}

Cevabı ve açıklamayı göster

Cevap: $0.90\text{ m/s}$

Cevap

The terminal velocity of the sphere is

0.90\text{ m/s}

.

At terminal velocity, the downward force of gravity (weight of the sphere) is balanced by the sum of two upward forces: the buoyant force (upthrust) and the viscous drag force given by Stokes' law (

F_v = 6\pi \eta r v_T

). Using the formula

v_T = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

with

r = 3.0 \times 10^{-3}\text{ m}

,

\rho_s - \rho_f = 4500\text{ kg/m}^3

,

\eta = 0.10\text{ Pa}\cdot\text{s}

, and

g = 10\text{ m/s}^2

yields

0.90\text{ m/s}

.

Adım Adım Çözüm

1

Convert given parameters to standard SI units

Radius

r = 3.0\text{ mm} = 3.0 \times 10^{-3}\text{ m}

, density of sphere

\rho_s = 5400\text{ kg/m}^3

, density of liquid

\rho_f = 900\text{ kg/m}^3

, viscosity

\eta = 0.10\text{ Pa}\cdot\text{s}

,

g = 10\text{ m/s}^2

.

Ensures dimensional consistency across all terms in the physical equations.

2

Apply the equilibrium condition at terminal velocity

At terminal velocity

v_T

, downward weight equals upward forces:

W = U + F_v

, where

W = \frac{4}{3}\pi r^3 \rho_s g

,

U = \frac{4}{3}\pi r^3 \rho_f g

, and

F_v = 6\pi \eta r v_T

.

Terminal velocity is reached when net acceleration is zero.

3

Rearrange Stokes' law formula for terminal velocity

v_T = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

Isolates the target unknown variable

v_T

.

4

Substitute the physical values and solve

v_T = \frac{2 \times (3.0 \times 10^{-3})^2 \times (5400 - 900) \times 10}{9 \times 0.10} = \frac{2 \times (9.0 \times 10^{-6}) \times 4500 \times 10}{0.90} = \frac{0.81}{0.90} = 0.90\text{ m/s}

.

Calculates the final quantitative answer.

Anahtar Kavram

Viscosity and Stokes' Law for terminal velocity of a sphere in a viscous fluid

Tahmini Süre:2m 0s

Soru 149Soru →

A simple pendulum of length $L$ and bob mass $m$ has a period of oscillation $T$ on the surface of the Earth. The length of the pendulum is increased by $44\%$ , its bob mass is doubled to $2m$ , and the entire apparatus is transported to a planet where the acceleration due to gravity is $36\%$ less than that on Earth. What is the new period of oscillation of the pendulum?

3.0T

1.5T

1.2T

0.67T

Cevabı ve açıklamayı göster

Cevap: $1.5T$

Cevap

1.5T

The period of a simple pendulum is determined by

T = 2\pi \sqrt{\frac{L}{g}}

. It is entirely independent of the mass of the bob. With a length increase of

44\%

(

L' = 1.44L

) and a gravity reduction of

36\%

(

g' = 0.64g

), the new period becomes

\sqrt{\frac{1.44}{0.64}}T = \frac{1.2}{0.8}T = 1.5T

.

Adım Adım Çözüm

1

State the equation for the period of a simple pendulum

T = 2\pi \sqrt{\frac{L}{g}}

The period depends solely on length

L

and acceleration due to gravity

g

, and is independent of mass

m

.

2

Determine the new length and new acceleration due to gravity

L' = 1.44L

and

g' = 0.64g

A

44\%

increase in length yields

1 + 0.44 = 1.44

, while a

36\%

decrease in gravity yields

1 - 0.36 = 0.64

.

3

Calculate the factor of change in the period

\frac{T'}{T} = \sqrt{\frac{L'}{L} \times \frac{g}{g'}} = \sqrt{\frac{1.44}{0.64}} = \sqrt{\frac{144}{64}} = \frac{12}{8} = 1.5

Substituting the relative changes into the pendulum period ratio yields the scaling factor.

4

Express the new period in terms of

T

T' = 1.5T

Multiplying the initial period by the calculated scaling factor gives the final period.

Anahtar Kavram

Mass independence and parametric scaling of simple pendulum period in Simple Harmonic Motion

Soru 150Soru →

A solid sphere of mass $0.60\text{ kg}$ and volume $2.0 \times 10^{-4}\text{ m}^3$ is released from rest in a tall vessel filled with a viscous liquid of density $1000\text{ kg/m}^3$ . As the sphere falls, it eventually reaches a constant terminal velocity of $4.0\text{ m/s}$ . Assuming that the viscous drag force is directly proportional to the speed of the sphere, what is the magnitude of the viscous drag force acting on the sphere when its speed is $1.5\text{ m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

0.75\text{ N}

1.50\text{ N}

2.25\text{ N}

2.50\text{ N}

Cevabı ve açıklamayı göster

Cevap: $1.50\text{ N}$

Cevap

1.50\text{ N}

At terminal velocity, the sphere is in translational equilibrium under three forces: downward weight (

6.0\text{ N}

), upward upthrust (

2.0\text{ N}

), and upward viscous drag (

4.0\text{ N}

). Since viscous drag is directly proportional to velocity (

F_v = k v

), the proportionality constant

k

is

1.0\text{ N}\cdot\text{s/m}

. Therefore, at

1.5\text{ m/s}

, the viscous force is

1.0 \times 1.5 = 1.50\text{ N}

.

Adım Adım Çözüm

1

Calculate the downward gravitational force (weight) acting on the sphere.

W = m \times g = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force pulling the sphere downward.

2

Calculate the buoyant force (upthrust) exerted by the liquid using Archimedes' principle.

U = \rho_l \times V \times g = 1000\text{ kg/m}^3 \times (2.0 \times 10^{-4}\text{ m}^3) \times 10\text{ m/s}^2 = 2.0\text{ N}

Upthrust equals the weight of the fluid displaced by the submerged sphere.

3

Determine the viscous drag force at terminal velocity (

v_t = 4.0\text{ m/s}

) using equilibrium of forces.

F_v(v_t) = W - U = 6.0\text{ N} - 2.0\text{ N} = 4.0\text{ N}

At terminal velocity, the net acceleration is zero, so downward weight is balanced by upward upthrust and viscous drag.

4

Find the constant of proportionality

k

for viscous drag (

F_v = k v

).

k = \frac{F_v(v_t)}{v_t} = \frac{4.0\text{ N}}{4.0\text{ m/s}} = 1.0\text{ N}\cdot\text{s/m}

Viscous force is given as directly proportional to speed.

5

Calculate the viscous drag force at a speed of

1.5\text{ m/s}

.

F_v(1.5) = k \times 1.5\text{ m/s} = 1.0\text{ N}\cdot\text{s/m} \times 1.5\text{ m/s} = 1.50\text{ N}

Applying the constant

k

to the specified speed.

Anahtar Kavram

Terminal velocity in viscous fluids and Archimedes' Principle

Tahmini Süre:2m 30s

Soru 151Soru →

A non-uniform wooden pole of length $6.0\text{ m}$ and weight $150\text{ N}$ is balanced horizontally on a pivot placed $2.4\text{ m}$ from its heavy end $P$ . The system achieves rotational equilibrium when a load of $50\text{ N}$ is hung directly from end $P$ . What is the distance of the center of gravity of the pole from end $P$ ?

Cevabı ve açıklamayı göster

Cevap: 3.2

Cevap

The distance of the center of gravity of the pole from end

P

is

3.2\text{ m}

.

Taking moments about the pivot at

2.4\text{ m}

from end

P

, the counter-clockwise moment created by the

50\text{ N}

load (

50\text{ N} \times 2.4\text{ m} = 120\text{ N}\cdot\text{m}

) must balance the clockwise moment created by the

150\text{ N}

weight of the pole acting at its center of gravity (

150\text{ N} \times (d - 2.4\text{ m})

). Equating these gives

120 = 150(d - 2.4)

, leading to

d - 2.4 = 0.8\text{ m}

, so

d = 3.2\text{ m}

.

Adım Adım Çözüm

1

Identify force positions relative to the pivot

The

50\text{ N}

load is

2.4\text{ m}

to the left of the pivot. The

150\text{ N}

weight acts at the center of gravity, which is

(d - 2.4\text{ m})

to the right of the pivot.

Moments are evaluated relative to the fulcrum to eliminate the unknown normal reaction force at the pivot.

2

Apply the Principle of Moments

Anti-clockwise moment =

50 \times 2.4 = 120\text{ N}\cdot\text{m}

. Clockwise moment =

150 \times (d - 2.4)

. Setting them equal:

120 = 150(d - 2.4)

.

For a body in rotational equilibrium, the total clockwise moment about any pivot equals the total anti-clockwise moment.

3

Solve the equation for distance

d

d - 2.4 = 0.8 \implies d = 3.2\text{ m}

.

Adding the displacement from the pivot (

0.8\text{ m}

) to the pivot position from end

P

(

2.4\text{ m}

) yields the position of the center of gravity from end

P

.

Anahtar Kavram

Rotational equilibrium and Principle of Moments for non-uniform rigid bodies

Tahmini Süre:1m 30s

Soru 152Soru →

A particle of mass $0.50\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.06\text{ m}$ , its speed is $0.80\text{ m/s}$ and its potential energy is $0.09\text{ J}$ . What is the magnitude of the maximum acceleration of the particle in $\text{m/s}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The magnitude of the maximum acceleration of the particle is

10\text{ m/s}^2

.

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

, the angular frequency

\omega

is

10\text{ rad/s}

. Using

v = \omega\sqrt{A^2 - x^2}

, the amplitude

A

is

0.10\text{ m}

. Substituting these values into

a_{\max} = \omega^2 A

yields

10\text{ m/s}^2

.

Adım Adım Çözüm

1

Find angular frequency

\omega

from potential energy.

\omega = 10\text{ rad/s}

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

:

0.09 = \frac{1}{2}(0.50)\omega^2 (0.06)^2 \implies 0.09 = 0.0009 \omega^2 \implies \omega^2 = 100 \implies \omega = 10\text{ rad/s}

.

2

Find amplitude

A

from speed.

A = 0.10\text{ m}

Using speed

v = \omega\sqrt{A^2 - x^2}

:

0.80 = 10\sqrt{A^2 - 0.06^2} \implies 0.08 = \sqrt{A^2 - 0.0036} \implies 0.0064 = A^2 - 0.0036 \implies A^2 = 0.0100 \implies A = 0.10\text{ m}

.

3

Calculate maximum acceleration.

a_{\max} = 10\text{ m/s}^2

Using maximum acceleration formula

a_{\max} = \omega^2 A

:

a_{\max} = 100 \times 0.10 = 10\text{ m/s}^2

.

Anahtar Kavram

Simple Harmonic Motion Energy and Kinematic Relations

Soru 153Soru →

A solid alloy specimen weighs $240\text{ g}$ in air, $180\text{ g}$ when completely immersed in water, and $195\text{ g}$ when completely immersed in an unknown liquid $X$ . What is the relative density of liquid $X$ ?

0.75

0.81

1.33

0.25

Cevabı ve açıklamayı göster

Cevap: $0.75$

Cevap

The relative density of liquid X is

0.75

.

According to Archimedes' principle, relative density of a liquid is the ratio of the upthrust experienced by a solid in that liquid to the upthrust experienced by the same solid in water. Upthrust in water is

240\text{ g} - 180\text{ g} = 60\text{ g}

, and upthrust in liquid X is

240\text{ g} - 195\text{ g} = 45\text{ g}

. Dividing

45

by

60

gives

0.75

.

Adım Adım Çözüm

1

Calculate the upthrust in water

240\text{ g} - 180\text{ g} = 60\text{ g}

By Archimedes' principle, upthrust in water equals the mass loss when immersed in water.

2

Calculate the upthrust in liquid X

240\text{ g} - 195\text{ g} = 45\text{ g}

Upthrust in liquid X equals the mass loss when immersed in liquid X.

3

Determine the relative density of liquid X

\text{Relative Density} = \frac{\text{Upthrust in liquid X}}{\text{Upthrust in water}} = \frac{45\text{ g}}{60\text{ g}} = 0.75

Relative density of a liquid is defined as the ratio of the weight of a given volume of the liquid to the weight of an equal volume of water.

Anahtar Kavram

Archimedes' Principle and Relative Density of Liquids

Tahmini Süre:1m 30s

Soru 154Soru →

A $0.40\text{ kg}$ mass attached to a light helical spring undergoes simple harmonic motion on a smooth horizontal surface. If the force constant of the spring is $160\text{ N/m}$ and the amplitude of oscillation is $0.05\text{ m}$ , what is the maximum speed of the mass?

1.0\text{ m/s}

20.0\text{ m/s}

0.40\text{ m/s}

4.00\text{ m/s}

Cevabı ve açıklamayı göster

Cevap: $1.0\text{ m/s}$

Cevap

The maximum speed of the mass is

1.0\text{ m/s}

.

The angular frequency of the mass-spring system is calculated using $\omega = \sqrt{k/m} = \sqrt{160/0.40} = 20\text{ rad/s}$. Multiplying this by the amplitude $A = 0.05\text{ m}$ yields a maximum speed of $v_{\text{max}} = 1.0\text{ m/s}$.

Adım Adım Çözüm

1

Calculate the angular frequency ($\omega$) of the mass-spring system.

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{160\text{ N/m}}{0.40\text{ kg}}} = \sqrt{400} = 20\text{ rad/s}$

The angular frequency of a spring-mass oscillator depends on the spring constant and the mass.

2

Determine the maximum speed ($v_{\text{max}}$) using the amplitude.

$v_{\text{max}} = \omega A = 20\text{ rad/s} \times 0.05\text{ m} = 1.0\text{ m/s}$

In simple harmonic motion, maximum speed occurs at the equilibrium position and equals the product of angular frequency and amplitude.

Anahtar Kavram

Maximum velocity in simple harmonic motion for a mass-spring system

Soru 155Soru →

A solid block of wood has a mass of $0.60\text{ kg}$ . When placed in a vessel of water, it floats freely on the surface. What is the magnitude of the upthrust exerted by the water on the block? (Take $g = 10\text{ m/s}^2$ )

6.0\text{ N}

0.60\text{ N}

0.0\text{ N}

60\text{ N}

Cevabı ve açıklamayı göster

Cevap: $6.0\text{ N}$

Cevap

The upthrust exerted by the water on the floating block is

6.0\text{ N}

.

According to the Law of Flotation, a body floating freely in a fluid displaces a weight of fluid equal to its own weight. Therefore, the upward force (upthrust) exerted by the fluid is equal to the weight of the block:

U = mg = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

.

Adım Adım Çözüm

1

Determine the weight of the floating block in air.

W = mg = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force of gravity acting on the block's mass.

2

Apply the Law of Flotation to find the upthrust.

Upthrust

U = W = 6.0\text{ N}

A freely floating body displaces a volume of fluid whose weight is equal to the total weight of the body.

Anahtar Kavram

Law of Flotation and Archimedes' Principle

Tahmini Süre:45s

Soru 156Soru →

A solid wooden block of density $600\text{ kg/m}^3$ and volume $4.0 \times 10^{-3}\text{ m}^3$ floats in water of density $1000\text{ kg/m}^3$ . A metal block is placed on top of the wooden block so that the wooden block is just completely submerged while the metal block remains entirely above the water surface. What is the mass of the metal block? (Take $g = 10\text{ m/s}^2$ )

1.6\text{ kg}

2.4\text{ kg}

4.0\text{ kg}

6.4\text{ kg}

Cevabı ve açıklamayı göster

Cevap: $1.6\text{ kg}$

Cevap

1.6\text{ kg}

When the wooden block is completely submerged, it displaces

4.0 \times 10^{-3}\text{ m}^3

of water, creating an upthrust of

40\text{ N}

. The weight of the wooden block is

24\text{ N}

. For equilibrium, the total downward weight must equal the upthrust (

24\text{ N} + W_{\text{metal}} = 40\text{ N}

), which gives

W_{\text{metal}} = 16\text{ N}

and a corresponding mass of

1.6\text{ kg}

.

Adım Adım Çözüm

1

Calculate the mass and weight of the wooden block.

m_{\text{wood}} = \rho_{\text{wood}} \times V_{\text{wood}} = 600\text{ kg/m}^3 \times 4.0 \times 10^{-3}\text{ m}^3 = 2.4\text{ kg}

, so

W_{\text{wood}} = 2.4 \times 10 = 24\text{ N}

.

The weight of the wood contributes to the total downward force of the floating system.

2

Calculate the total upthrust exerted by the water when the wooden block is completely submerged.

U = \rho_{\text{water}} \times V_{\text{wood}} \times g = 1000\text{ kg/m}^3 \times 4.0 \times 10^{-3}\text{ m}^3 \times 10\text{ m/s}^2 = 40\text{ N}

.

By Archimedes' principle, upthrust equals the weight of the displaced liquid.

3

Apply the law of flotation to solve for the weight and mass of the metal block.

W_{\text{wood}} + W_{\text{metal}} = U \implies 24\text{ N} + W_{\text{metal}} = 40\text{ N} \implies W_{\text{metal}} = 16\text{ N}

. Thus,

m_{\text{metal}} = \frac{16\text{ N}}{10\text{ m/s}^2} = 1.6\text{ kg}

.

For the system to float in equilibrium just submerged, total downward weight must equal total upward buoyant force.

Anahtar Kavram

Archimedes' Principle and Law of Flotation

Tahmini Süre:2m 0s

Soru 157Soru →

A vertical light spring stretches by $0.10\text{ m}$ when a block is suspended from it in equilibrium. The block is then pulled down an additional $0.05\text{ m}$ from its equilibrium position and released from rest to undergo simple harmonic motion. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the block when it is at a displacement of $0.03\text{ m}$ from its equilibrium position?

0.40\text{ m/s}

0.50\text{ m/s}

0.30\text{ m/s}

0.04\text{ m/s}

Cevabı ve açıklamayı göster

Cevap: $0.40\text{ m/s}$

Cevap

0.40\text{ m/s}

The system's angular frequency

\omega

is determined by the equilibrium extension

e

using

\omega = \sqrt{\frac{g}{e}} = 10\text{ rad/s}

. Combining this with the amplitude

A = 0.05\text{ m}

in the SHM speed relation

v = \omega \sqrt{A^2 - y^2}

at

y = 0.03\text{ m}

yields

v = 10 \sqrt{(0.05)^2 - (0.03)^2} = 0.40\text{ m/s}

.

Adım Adım Çözüm

1

Calculate the angular frequency of the mass-spring system using static equilibrium conditions.

\omega = 10\text{ rad/s}

At equilibrium,

mg = ke \implies \frac{k}{m} = \frac{g}{e}

. Therefore,

\omega = \sqrt{\frac{g}{e}} = \sqrt{\frac{10}{0.10}} = 10\text{ rad/s}

.

2

Identify the amplitude of simple harmonic motion.

A = 0.05\text{ m}

The initial displacement from the equilibrium position when released from rest defines the amplitude of oscillation.

3

Apply the SHM velocity-displacement formula

v = \omega \sqrt{A^2 - y^2}

at

y = 0.03\text{ m}

.

v = 0.40\text{ m/s}

v = 10 \times \sqrt{(0.05)^2 - (0.03)^2} = 10 \times \sqrt{0.0025 - 0.0009} = 10 \times 0.04 = 0.40\text{ m/s}

.

Anahtar Kavram

Relating static extension to angular frequency and calculating instantaneous speed in Simple Harmonic Motion

Soru 158Soru →

A particle of mass $0.20\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.03\text{ m}$ , its speed is $0.16\text{ m/s}$ . When its displacement is $0.04\text{ m}$ , its speed is $0.12\text{ m/s}$ . What is the total mechanical energy of the particle in millijoules ( $\text{mJ}$ )?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The total mechanical energy of the particle is

4\text{ mJ}

.

Using the relation

v^2 = \omega^2(A^2 - x^2)

for the two given state points

(0.03\text{ m}, 0.16\text{ m/s})

and

(0.04\text{ m}, 0.12\text{ m/s})

forms a set of simultaneous equations. Subtracting them yields

\omega^2 = 16\text{ rad}^2/\text{s}^2

, leading to

A^2 = 0.0025\text{ m}^2

. Substituting these values into

E = \frac{1}{2}m\omega^2 A^2

gives

E = 0.004\text{ J}

, which converts to

4\text{ mJ}

.

Adım Adım Çözüm

1

Set up kinematic equations for both displacement points using

v^2 = \omega^2(A^2 - x^2)

.

0.0256 = \omega^2(A^2 - 0.0009)

and

0.0144 = \omega^2(A^2 - 0.0016)

.

The equation relates linear speed, angular frequency, amplitude, and instantaneous displacement in SHM.

2

Subtract the two simultaneous equations to eliminate

A^2

and find

\omega^2

.

0.0112 = 0.0007\omega^2 \implies \omega^2 = 16\text{ rad}^2/\text{s}^2

.

Eliminating amplitude isolates the angular frequency squared.

3

Determine

A^2

by substituting

\omega^2 = 16

back into one of the state equations.

A^2 = 0.0025\text{ m}^2 \implies A = 0.05\text{ m}

.

Amplitude is required to calculate the maximum potential or total mechanical energy.

4

Calculate total mechanical energy

E = \frac{1}{2}m\omega^2 A^2

and convert to millijoules.

E = \frac{1}{2} \times 0.20 \times 16 \times 0.0025 = 0.004\text{ J} = 4\text{ mJ}

.

Total energy in SHM is constant and proportional to mass, square of angular frequency, and square of amplitude.

Anahtar Kavram

Conservation of energy and phase-space relationship between velocity and displacement in simple harmonic motion.

Soru 159Soru →

A small spherical lead shot of radius $3.0 \times 10^{-3}\text{ m}$ and density $8000\text{ kg/m}^3$ falls vertically through a viscous oil of density $800\text{ kg/m}^3$ and dynamic viscosity $0.18\text{ Pa}\cdot\text{s}$ . What is the magnitude of the terminal velocity of the sphere in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

Cevabı ve açıklamayı göster

Cevap: 0.8

Cevap

The terminal velocity of the lead shot is

0.8\text{ m/s}

.

At terminal velocity, acceleration is zero because the upward forces (viscous drag

6\pi \eta r v_t

plus buoyant upthrust

\frac{4}{3}\pi r^3 \rho_f g

) completely balance the downward weight of the sphere (

\frac{4}{3}\pi r^3 \rho_s g

). Solving for velocity gives

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta} = 0.8\text{ m/s}

.

Adım Adım Çözüm

1

Formulate the force balance equation at terminal velocity.

Weight (

W

) = Upthrust (

U

) + Viscous drag (

F_v

), which simplifies to

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

.

When terminal velocity is attained, the net acceleration of the sphere is zero, so upward forces balance downward force.

2

Substitute the physical parameters into the terminal velocity formula.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 800) \times 10}{9 \times 0.18} = 0.8\text{ m/s}

.

Direct calculation using Stokes' law and Archimedes' principle.

Anahtar Kavram

Terminal Velocity and Stokes' Law in a Viscous Medium

Soru 160Soru →

A uniform horizontal rod $AB$ of length $2.0\text{ m}$ and mass $6.0\text{ kg}$ is suspended horizontally by two light vertical strings attached at end $A$ and at a point $C$ located $0.5\text{ m}$ from end $B$ . Taking $g = 10\text{ m/s}^2$ , what is the tension in the string at point $C$ in Newtons?

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The tension in the string at point C is 40 N.

The weight of the rod (

60\text{ N}

) acts at its midpoint (

1.0\text{ m}

from end

A

). Point

C

is located

1.5\text{ m}

from end

A

. Taking moments about end

A

gives

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

, which evaluates to

T_C = 40\text{ N}

.

Adım Adım Çözüm

1

Calculate total weight and identify the center of gravity position

Weight

W = 60\text{ N}

acting at

1.0\text{ m}

from end

A

For a uniform rod, the weight acts vertically downwards at its geometric center.

2

Set up the moment equilibrium equation taking end A as pivot

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

Taking moments about point

A

eliminates the force at

A

and equates clockwise moment from weight to counter-clockwise moment from tension at

C

.

3

Solve for the tension force at point C

T_C = 40\text{ N}

Dividing the total moment of

60\text{ N}\cdot\text{m}

by the moment arm of

1.5\text{ m}

yields

40\text{ N}

.

Anahtar Kavram

Principle of Moments and Rotational Equilibrium

Tahmini Süre:1m 30s

Mechanics

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