Mechanics Alıştırma Soruları — JAMB UTME — Sayfa 10

Soru 181Soru →

A simple machine lifts a load of $200\text{ N}$ when an effort of $50\text{ N}$ is applied to it. What is the mechanical advantage of the machine?

Cevabı ve açıklamayı göster

Cevap: 4; 4.0

Cevap

The mechanical advantage of the machine is

4

.

Mechanical Advantage (MA) is the ratio of the force exerted by the machine (load) to the force applied to the machine (effort). Dividing

200\text{ N}

by

50\text{ N}

yields a dimensionless mechanical advantage of

4

.

Adım Adım Çözüm

1

Identify the given values from the problem statement

Load (

L

) =

200\text{ N}

, Effort (

E

) =

50\text{ N}

Mechanical advantage is defined as the ratio of load to effort.

2

Apply the formula for Mechanical Advantage (MA)

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{200\text{ N}}{50\text{ N}} = 4

Dividing the output force by the input force gives the force amplification factor of the machine.

Anahtar Kavram

Mechanical Advantage of Simple Machines

Tahmini Süre:45s

Soru 182Soru →

A simple machine with a velocity ratio of $5$ requires an effort of $200\text{ N}$ to raise a load of $800\text{ N}$ . What is the efficiency of the machine in percentage?

Cevabı ve açıklamayı göster

Cevap: 80%; 80; 80 percent; 80 %

Cevap

The efficiency of the machine is

80\%

.

The mechanical advantage is calculated by dividing the load (

800\text{ N}

) by the effort (

200\text{ N}

), yielding

\text{MA} = 4

. Dividing the mechanical advantage by the velocity ratio (

5

) and multiplying by

100\%

gives an efficiency of

80\%

.

Adım Adım Çözüm

1

Calculate the Mechanical Advantage (MA) of the machine

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{800\text{ N}}{200\text{ N}} = 4

Mechanical Advantage is defined as the ratio of the force overcome (load) to the force applied (effort).

2

Calculate the Efficiency using Mechanical Advantage and Velocity Ratio

\text{Efficiency} = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100\% = \left(\frac{4}{5}\right) \times 100\% = 80\%

Efficiency is the ratio of Mechanical Advantage to Velocity Ratio expressed as a percentage.

Anahtar Kavram

Efficiency of a Simple Machine

Tahmini Süre:1m 30s

Soru 183Soru →

A uniform rigid beam $MN$ of length $2.0\text{ m}$ and weight $80\text{ N}$ is smoothly pivoted at end $M$ . A vertical load of $120\text{ N}$ is hung at a distance of $1.5\text{ m}$ from $M$ . The beam is held horizontally in equilibrium by a cable attached at end $N$ pulling upward at an angle of $30^\circ$ to the horizontal beam. What is the magnitude of the tension in the cable?

100\text{ N}

130\text{ N}

260\text{ N}

340\text{ N}

Cevabı ve açıklamayı göster

Cevap: $260\text{ N}$

Cevap

The magnitude of the tension in the cable is

260\text{ N}

.

For rotational equilibrium about pivot

M

, the sum of clockwise moments must equal the counterclockwise moment. Clockwise moment from the beam's weight and load is

(80\text{ N} \times 1.0\text{ m}) + (120\text{ N} \times 1.5\text{ m}) = 260\text{ N}\cdot\text{m}

. Counterclockwise moment from the cable tension is

T \times 2.0\text{ m} \times \sin(30^\circ) = 1.0 T\text{ N}\cdot\text{m}

. Equating the two gives

T = 260\text{ N}

.

Adım Adım Çözüm

1

Identify the positions and lines of action of all forces relative to pivot

M

.

The weight of the uniform beam (

80\text{ N}

) acts at its center of gravity (

1.0\text{ m}

from

M

). The suspended load (

120\text{ N}

) acts at

1.5\text{ m}

from

M

. Cable tension

T

acts at

2.0\text{ m}

from

M

at an angle of

30^\circ

to the beam.

Rotational equilibrium requires evaluating moments created by all forces about the pivot point.

2

Calculate the sum of clockwise moments about pivot

M

.

\sum \tau_{\text{clockwise}} = (80\text{ N} \times 1.0\text{ m}) + (120\text{ N} \times 1.5\text{ m}) = 80\text{ N}\cdot\text{m} + 180\text{ N}\cdot\text{m} = 260\text{ N}\cdot\text{m}

.

Both downward forces exert clockwise turning effects about pivot

M

.

3

Determine the counterclockwise moment exerted by the inclined cable tension

T

.

\tau_{\text{counterclockwise}} = T \times d \sin\theta = T \times 2.0\text{ m} \times \sin(30^\circ) = 1.0 T\text{ N}\cdot\text{m}.

Only the component of tension perpendicular to the beam (

T \sin 30^\circ

) produces a moment about the pivot.

4

Apply the Principle of Moments to calculate tension

T

.

1.0 T = 260 \implies T = 260\text{ N}.

For the beam to remain horizontally balanced, total clockwise moment must equal total counterclockwise moment.

Anahtar Kavram

Principle of Moments and Rotational Equilibrium with Inclined Forces

Soru 184Soru →

An object of mass $4\text{ kg}$ is released from rest at a height of $5\text{ m}$ above the ground. Neglecting air resistance and taking $g = 10\text{ m s}^{-2}$ , what is its kinetic energy just before striking the ground?

200\text{ J}

100\text{ J}

40\text{ J}

20\text{ J}

Cevabı ve açıklamayı göster

Cevap: $200\text{ J}$

Cevap

The kinetic energy of the object just before striking the ground is

200\text{ J}

.

The total mechanical energy is conserved during free fall. The initial gravitational potential energy

E_p = mgh = 4 \times 10 \times 5 = 200\text{ J}

is completely converted into kinetic energy

E_k

just before impact, making

200\text{ J}

the correct answer.

Adım Adım Çözüm

1

Calculate the initial potential energy at maximum height

E_p = mgh = 4\text{ kg} \times 10\text{ m s}^{-2} \times 5\text{ m} = 200\text{ J}

At the top of the fall, all mechanical energy is stored as gravitational potential energy.

2

Apply the principle of conservation of mechanical energy

E_k = E_p = 200\text{ J}

In the absence of resistive forces such as air drag, potential energy lost converts completely into kinetic energy gained.

Anahtar Kavram

Conservation of Mechanical Energy

Tahmini Süre:45s

Soru 185Soru →

A body of mass $0.50\text{ kg}$ connected to a light helical spring of force constant $32\text{ N/m}$ executes simple harmonic motion on a smooth horizontal surface. If the total mechanical energy of the oscillating system is $0.16\text{ J}$ , what is the maximum speed of the body in $\text{m/s}$ ?

Cevabı ve açıklamayı göster

Cevap: 0.8

Cevap

The maximum speed of the body is

0.8\text{ m/s}

.

The total mechanical energy in simple harmonic motion is equal to the maximum kinetic energy at the equilibrium position:

E = \frac{1}{2} m v_{\text{max}}^2

. Substituting

E = 0.16\text{ J}

and

m = 0.50\text{ kg}

yields

0.16 = 0.25 v_{\text{max}}^2

, so

v_{\text{max}}^2 = 0.64

and

v_{\text{max}} = 0.8\text{ m/s}

.

Adım Adım Çözüm

1

Relate total energy to maximum kinetic energy

E = \frac{1}{2} m v_{\text{max}}^2

At the equilibrium position, potential energy is zero and total mechanical energy is entirely kinetic.

2

Substitute given values into the equation

0.16 = \frac{1}{2} (0.50) v_{\text{max}}^2

Mass

m = 0.50\text{ kg}

and total energy

E = 0.16\text{ J}

are provided.

3

Solve for the maximum speed

v_{\text{max}} = \sqrt{\frac{2 \times 0.16}{0.50}} = \sqrt{0.64} = 0.8\text{ m/s}

Solving for

v_{\text{max}}

gives

0.8\text{ m/s}

.

Anahtar Kavram

Conservation of Energy in Simple Harmonic Motion

Soru 186Soru →

A particle executing simple harmonic motion has a maximum speed of $3.0\text{ m/s}$ and a maximum acceleration of $12.0\text{ m/s}^2$ . What is the period of oscillation of the particle?

\frac{\pi}{4}\text{ s}

\frac{\pi}{2}\text{ s}

\pi\text{ s}

2\pi\text{ s}

Cevabı ve açıklamayı göster

Cevap: $\frac{\pi}{2}\text{ s}$

Cevap

The period of oscillation of the particle is

\frac{\pi}{2}\text{ s}

.

For simple harmonic motion, maximum speed is

v_{\text{max}} = \omega A

and maximum acceleration is

a_{\text{max}} = \omega^2 A

. Dividing the maximum acceleration by the maximum speed gives

\omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{12.0}{3.0} = 4.0\text{ rad/s}

. Using the formula for the period

T = \frac{2\pi}{\omega}

, we find

T = \frac{2\pi}{4.0} = \frac{\pi}{2}\text{ s}

.

Adım Adım Çözüm

1

Relate maximum speed and maximum acceleration to angular frequency

\omega = \frac{a_{\text{max}}}{v_{\text{max}}}

Since

v_{\text{max}} = \omega A

and

a_{\text{max}} = \omega^2 A

, dividing

a_{\text{max}}

by

v_{\text{max}}

eliminates the amplitude

A

and gives

\omega

.

2

Calculate the angular frequency

\omega

\omega = \frac{12.0\text{ m/s}^2}{3.0\text{ m/s}} = 4.0\text{ rad/s}

Substitute the given numerical values into the expression for angular frequency.

3

Calculate the period

T

T = \frac{2\pi}{\omega} = \frac{2\pi}{4.0} = \frac{\pi}{2}\text{ s}

The period of simple harmonic motion is related to angular frequency by

T = \frac{2\pi}{\omega}

.

Anahtar Kavram

Relationship between maximum velocity, maximum acceleration, angular frequency, and period in simple harmonic motion.

Tahmini Süre:1m 15s

Soru 187Soru →

An electric water pump with an efficiency of $80\%$ lifts $60\text{ kg}$ of water vertically through a height of $10\text{ m}$ in $1\text{ minute}$ . What is the input power of the pump in watts? (Take $g = 10\text{ m s}^{-2}$ )

Cevabı ve açıklamayı göster

Cevap: 125

Cevap

The input power of the pump is

125\text{ W}

.

The total gravitational potential energy gained by

60\text{ kg}

of water lifted

10\text{ m}

is

W = mgh = 60 \times 10 \times 10 = 6000\text{ J}

. Performed over

60\text{ seconds}

, the useful power output is

100\text{ W}

. Accounting for an efficiency of

80\%

(

0.80

), the required input power is

P_{\text{in}} = \frac{100\text{ W}}{0.80} = 125\text{ W}

.

Adım Adım Çözüm

1

Calculate the useful work done to lift the water

W = mgh = 60\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 6000\text{ J}

The useful work done equals the gravitational potential energy gained by the lifted mass of water.

2

Determine the useful output power of the pump

P_{\text{out}} = \frac{W}{t} = \frac{6000\text{ J}}{60\text{ s}} = 100\text{ W}

Power is the rate at which work is done, and

1\text{ minute}

must be converted to

60\text{ seconds}

.

3

Calculate the required input power using efficiency

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{100\text{ W}}{0.80} = 125\text{ W}

Efficiency is defined as

\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}

, so

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}}

.

Anahtar Kavram

Work, Power, and Efficiency of a Pump System

Tahmini Süre:1m 30s

Soru 188Soru →

In an isolated system where two colliding bodies of unequal mass undergo a perfectly elastic head-on collision, the body with the larger mass imparts a greater magnitude of impulse on the lighter body than the lighter body imparts on the heavier body.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is false. By Newton's Third Law, interaction forces are equal in magnitude and opposite in direction at all times, making the impulse imparted by each body on the other equal in magnitude.

The statement is false because Newton's Third Law dictates that the force exerted by object 1 on object 2 is equal in magnitude to the force exerted by object 2 on object 1 at every instant during contact. Integrating force over time yields equal magnitudes of impulse (

|J_1| = |J_2|

), independent of mass differences.

Adım Adım Çözüm

1

Apply Newton's Third Law to interaction forces.

F_{12} = -F_{21}

, where

F_{12}

is the force exerted on body 2 by body 1, and

F_{21}

is the force exerted on body 1 by body 2.

Forces between interacting objects always occur in equal and opposite action-reaction pairs regardless of mass.

2

Integrate both forces over the duration of collision

\Delta t

.

J_{12} = \int F_{12} \, dt = -\int F_{21} \, dt = -J_{21}

.

Impulse is defined as the time-integral of force.

3

Compare impulse magnitudes.

|J_{12}| = |J_{21}|

.

Taking the magnitude of both sides shows that both bodies experience equal magnitudes of impulse regardless of mass ratio or collision elasticity.

Anahtar Kavram

Newton's Third Law and Equality of Mutual Impulse

Soru 189Soru →

A spring with a stiffness constant of $200\text{ N m}^{-1}$ is compressed by $0.3\text{ m}$ on a frictionless horizontal table. A block of mass $0.5\text{ kg}$ is placed against the compressed spring. When the system is released from rest, all the stored elastic potential energy of the spring is transferred to the block. What is the speed of the block as it leaves the spring?

6.0\text{ m s}^{-1}

36.0\text{ m s}^{-1}

18.0\text{ m s}^{-1}

3.0\text{ m s}^{-1}

Cevabı ve açıklamayı göster

Cevap: $6.0\text{ m s}^{-1}$

Cevap

The speed of the block as it leaves the spring is

6.0\text{ m s}^{-1}

.

By the law of conservation of energy, the elastic potential energy stored in the spring when compressed by

x = 0.3\text{ m}

is

E_p = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.3)^2 = 9\text{ J}

. Upon release, this energy converts fully into the block's kinetic energy

E_k = \frac{1}{2}mv^2 = 9\text{ J}

. Substituting

m = 0.5\text{ kg}

gives

0.25 v^2 = 9

, so

v^2 = 36

and

v = 6.0\text{ m s}^{-1}

.

Adım Adım Çözüm

1

Calculate the elastic potential energy (

E_p

) stored in the compressed spring.

E_p = \frac{1}{2} k x^2 = \frac{1}{2} \times 200 \times (0.3)^2 = 100 \times 0.09 = 9.0\text{ J}

Energy stored in a compressed ideal spring is given by Hooke's law energy formula.

2

Apply the law of conservation of mechanical energy to find the kinetic energy (

E_k

) of the block.

E_k = E_p = 9.0\text{ J}

On a frictionless surface, all elastic potential energy converts entirely into translational kinetic energy.

3

Solve for the velocity (

v

) using the kinetic energy formula

E_k = \frac{1}{2} m v^2

.

9.0 = \frac{1}{2} (0.5) v^2 \implies 0.25 v^2 = 9.0 \implies v^2 = 36 \implies v = 6.0\text{ m s}^{-1}

Isolating

v

requires dividing by half the mass and taking the principal square root.

Anahtar Kavram

Conservation of Mechanical Energy (Elastic Potential Energy to Kinetic Energy)

Soru 190Soru →

A simple pendulum has a period of oscillation of $1.6\text{ s}$ on the surface of the Earth, where the acceleration due to gravity is $10.0\text{ m/s}^2$ . What is the period of oscillation of this pendulum when placed on a moon where the acceleration due to gravity is $2.5\text{ m/s}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 3.2

Cevap

The period of oscillation of the pendulum on the moon is

3.2\text{ s}

.

The period of a simple pendulum is given by

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

. Because the length

l

is constant, period is inversely proportional to the square root of acceleration due to gravity (

T \propto \frac{1}{\sqrt{g}}

). Reducing the local gravity from

10.0\text{ m/s}^2

to

2.5\text{ m/s}^2

decreases gravity by a factor of 4, which increases the period by a factor of

\sqrt{4} = 2

. Multiplying the initial period of

1.6\text{ s}

by 2 yields

3.2\text{ s}

.

Adım Adım Çözüm

1

Relate the period of oscillation of a simple pendulum to gravitational acceleration.

The period formula is

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

, showing that

T

is inversely proportional to

\sqrt{g}

.

The length of the pendulum

l

remains unchanged.

2

Formulate a ratio comparing the pendulum's period on the moon to its period on Earth.

\frac{T_{moon}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{moon}}}

Dividing the two equations cancels the constant terms

2\pi

and

\sqrt{l}

.

3

Substitute the known numerical values and solve for

T_{moon}

.

T_{moon} = 1.6 \times \sqrt{\frac{10.0}{2.5}} = 1.6 \times 2.0 = 3.2\text{ s}

The ratio of gravities is 4, whose square root is 2, doubling the initial period.

Anahtar Kavram

Dependence of Simple Pendulum Period on Gravitational Acceleration

Tahmini Süre:1m 30s

Soru 191Soru →

A particle executes simple harmonic motion along a straight line with an angular frequency of $6.0\text{ rad/s}$ . What is the magnitude of the acceleration of the particle, in $\text{m/s}^2$ , when its displacement from the mean position is $0.50\text{ m}$ ?

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

18.0 m/s^2

The magnitude of acceleration in simple harmonic motion is calculated using

a = \omega^2 x

. Substituting

\omega = 6.0\text{ rad/s}

and

x = 0.50\text{ m}

gives

a = (6.0)^2 \times 0.50 = 36 \times 0.50 = 18.0\text{ m/s}^2

.

Adım Adım Çözüm

1

Identify the formula relating acceleration to angular frequency and displacement in SHM.

a = \omega^2 x

In simple harmonic motion, the magnitude of acceleration is directly proportional to displacement from the equilibrium position.

2

Substitute the given physical values into the equation.

a = (6.0)^2 \times 0.50

The given values are angular frequency

\omega = 6.0\text{ rad/s}

and displacement

x = 0.50\text{ m}

.

3

Compute the numerical product.

a = 18.0\text{ m/s}^2

Squaring

6.0

yields

36

, and multiplying by

0.50

gives

18.0

.

Anahtar Kavram

Acceleration in Simple Harmonic Motion

Tahmini Süre:1m 0s

Soru 192Soru →

A light rigid rod $OP$ of length $1.2\text{ m}$ is pivoted smoothly at end $O$ . A vertical downward load of $30\text{ N}$ is suspended from end $P$ . An upward force $F$ is applied at the midpoint of the rod at an angle of $30^\circ$ to the horizontal rod to keep it in horizontal equilibrium. What is the magnitude of the force $F$ ?

60\text{ N}

120\text{ N}

240\text{ N}

40\text{ N}

Cevabı ve açıklamayı göster

Cevap: $120\text{ N}$

Cevap

The magnitude of the force

F

is

120\text{ N}

.

The correct answer is

120\text{ N}

. For rotational equilibrium about the pivot, the clockwise moment created by the suspended load (

30\text{ N} \times 1.2\text{ m} = 36\text{ N}\cdot\text{m}

) must equal the counterclockwise moment generated by force

F

. Because force

F

acts at an angle of

30^\circ

at the midpoint (

0.6\text{ m}

), its effective perpendicular component is

F \sin(30^\circ) = 0.5F

. Equating the moments gives

0.5F \times 0.6\text{ m} = 36\text{ N}\cdot\text{m}

, which yields

0.3F = 36

, resulting in

F = 120\text{ N}

.

Adım Adım Çözüm

1

Calculate the clockwise moment about the pivot

O

due to the load at end

P

.

\tau_{\text{clockwise}} = 30\text{ N} \times 1.2\text{ m} = 36\text{ N}\cdot\text{m}

The force of

30\text{ N}

acts vertically downwards at a perpendicular distance of

1.2\text{ m}

from the pivot.

2

Determine the perpendicular distance (or perpendicular component of force) for

F

applied at the midpoint.

Midpoint distance =

\frac{1.2\text{ m}}{2} = 0.6\text{ m}

; Perpendicular force component =

F \sin(30^\circ) = 0.5F

Only the component perpendicular to the line of action contributes to the moment about pivot

O

.

3

Set up the counterclockwise moment expression about pivot

O

.

\tau_{\text{counterclockwise}} = F \sin(30^\circ) \times 0.6\text{ m} = 0.3F\text{ N}\cdot\text{m}

The moment is the product of the perpendicular force component and the distance from the pivot.

4

Equate clockwise and counterclockwise moments to solve for

F

.

0.3F = 36 \implies F = \frac{36}{0.3} = 120\text{ N}

According to the principle of moments, total clockwise moments must equal total counterclockwise moments for rotational equilibrium.

Anahtar Kavram

Principle of Moments and Rotational Equilibrium

Tahmini Süre:1m 15s

Soru 193Soru →

Two satellites, $X$ and $Y$ , move in circular orbits around a planet with orbital radii of $r$ and $4r$ , respectively. If satellite $X$ travels at an orbital speed of $v$ , what is the orbital speed of satellite $Y$ ?

\frac{v}{2}

\frac{v}{4}

2v

4v

Cevabı ve açıklamayı göster

Cevap: $\frac{v}{2}$

Cevap

The orbital speed of satellite

Y

is

\frac{v}{2}

.

The orbital speed

v

of a satellite in a circular orbit of radius

r

around a planet of mass

M

is given by

v = \sqrt{\frac{GM}{r}}

. When the radius increases by a factor of 4 from

r

to

4r

, the new speed becomes

v_Y = \sqrt{\frac{GM}{4r}} = \frac{1}{2} \sqrt{\frac{GM}{r}} = \frac{v}{2}

.

Adım Adım Çözüm

1

Write the formula for orbital speed of a satellite

v = \sqrt{\frac{GM}{r}}

Orbital speed depends on the gravitational mass

M

of the central body and the orbital radius

r

.

2

Set up the ratio between the orbital speeds of satellites

Y

and

X

\frac{v_Y}{v_X} = \sqrt{\frac{r_X}{r_Y}}

Since

G

and

M

are constant, orbital speed is inversely proportional to

\sqrt{r}

.

3

Substitute the given values

r_X = r

,

r_Y = 4r

, and

v_X = v

v_Y = v \sqrt{\frac{r}{4r}} = v \sqrt{\frac{1}{4}} = \frac{v}{2}

Evaluating the square root yields a factor of

\frac{1}{2}

.

Anahtar Kavram

Orbital Speed of a Satellite

Tahmini Süre:1m 0s

Soru 194Soru →

A simple pendulum of fixed length $l$ carries a bob of mass $m$ and oscillates with a period $T$ . If the bob is replaced by another bob of mass $2m$ while maintaining the exact same string length, what is the new period of oscillation?

2T

T

\sqrt{2}T

\frac{T}{2}

Cevabı ve açıklamayı göster

Cevap: $T$

Cevap

The new period of oscillation remains

T

.

The period of oscillation for a simple pendulum undergoing small displacement SHM is given by

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

. The mass of the bob

m

is not a parameter in this equation. Therefore, altering the mass while keeping length

l

constant leaves the period unchanged as

T

.

Adım Adım Çözüm

1

Identify the governing equation for the period of a simple pendulum executing simple harmonic motion.

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

To analyze which physical parameters determine the period of the simple pendulum.

2

Check for mass dependence in the formula.

The mass variable

m

does not appear in

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

.

The restoring gravitational force and inertia both scale linearly with mass, causing mass to cancel out entirely.

3

Determine the period after doubling the mass at constant length.

The new period is equal to

T

.

Since length

l

and acceleration due to gravity

g

are constant, changing the mass from

m

to

2m

has no effect on period.

Anahtar Kavram

Independence of Simple Pendulum Period from Bob Mass

Soru 195Soru →

An object of mass $m$ is projected vertically upwards from the surface of a spherical planet of radius $R$ and mass $M$ with an initial speed equal to half of the planet's escape velocity. Neglecting atmospheric friction, what maximum height above the planet's surface will the object reach?

\frac{R}{4}

\frac{R}{3}

\frac{R}{2}

R

Cevabı ve açıklamayı göster

Cevap: $\frac{R}{3}$

Cevap

The maximum height above the planet's surface reached by the object is

\frac{R}{3}

.

By mechanical energy conservation, the initial total energy at launch equals the potential energy at maximum height where velocity is zero. The launch velocity

v = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}

gives an initial kinetic energy of

\frac{GMm}{4R}

. Combined with initial potential energy

-\frac{GMm}{R}

, total energy is

-\frac{3GMm}{4R}

. Equating this to

-\frac{GMm}{R+h}

gives

R+h = \frac{4}{3}R

, yielding a height above the surface of

h = \frac{R}{3}

.

Adım Adım Çözüm

1

Express the initial speed in terms of gravitational constant

G

, mass

M

, and radius

R

.

The escape velocity is

v_e = \sqrt{\frac{2GM}{R}}

. Thus, launch speed

v = \frac{1}{2}v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}

.

Escape velocity is defined as the minimum speed needed to escape the gravitational field.

2

Calculate the initial total mechanical energy at the planet's surface.

Initial kinetic energy

K_i = \frac{1}{2}m v^2 = \frac{1}{2}m\left(\frac{2GM}{4R}\right) = \frac{GMm}{4R}

. Surface potential energy

U_i = -\frac{GMm}{R}

. Total energy

E_i = K_i + U_i = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}

.

Total mechanical energy is the sum of kinetic energy and gravitational potential energy.

3

Apply energy conservation to find the maximum distance from the planet's center.

At maximum height

h

, speed is zero (

K_f = 0

), so distance from center is

r = R + h

. Energy

E_f = -\frac{GMm}{R+h}

. Equating

E_i = E_f

gives

-\frac{3GMm}{4R} = -\frac{GMm}{R+h} \implies R+h = \frac{4}{3}R \implies h = \frac{R}{3}

.

Mechanical energy is conserved in a central gravitational force field.

Anahtar Kavram

Conservation of Mechanical Energy in a Gravitational Field

Tahmini Süre:1m 30s

Soru 196Soru →

A gear system consists of a driving gear with $12$ teeth and a driven gear with $48$ teeth. If an effort of $40\text{ N}$ applied to the driving gear overcomes a load of $120\text{ N}$ on the driven gear, what is the efficiency of the machine?

Cevabı ve açıklamayı göster

Cevap: 75%; 75 %; 75 percent; 75

Cevap

The efficiency of the gear system is 75%.

The velocity ratio is determined by dividing the number of teeth on the driven gear (48) by the number of teeth on the driving gear (12), giving a VR of 4. The mechanical advantage is the ratio of load (120 N) to effort (40 N), giving an MA of 3. Dividing the mechanical advantage by the velocity ratio and multiplying by 100 yields an efficiency of 75%.

Adım Adım Çözüm

1

Calculate the velocity ratio (VR) of the gear system.

VR = 4

For a gear system, the velocity ratio is the ratio of the number of teeth on the driven gear to the number of teeth on the driving gear:

\text{VR} = \frac{N_{\text{driven}}}{N_{\text{driving}}} = \frac{48}{12} = 4

.

2

Calculate the mechanical advantage (MA) of the gear system.

MA = 3

Mechanical advantage is the ratio of the load to the effort:

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{120\text{ N}}{40\text{ N}} = 3

.

3

Calculate the efficiency of the gear system.

Efficiency = 75%

Efficiency is given by the formula:

\text{Efficiency} = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{3}{4} \times 100\% = 75\%

.

Anahtar Kavram

Efficiency, Mechanical Advantage, and Velocity Ratio of a Gear System

Tahmini Süre:1m 30s

Soru 197Soru →

A particle of mass $0.20\text{ kg}$ executes simple harmonic motion with an amplitude of $0.05\text{ m}$ and a period of oscillation of $0.20\pi\text{ s}$ . What is the maximum kinetic energy of the particle in joules?

Cevabı ve açıklamayı göster

Cevap: 0.025

Cevap

The maximum kinetic energy of the particle is

0.025\text{ J}

.

The maximum kinetic energy occurs at the equilibrium position where speed reaches its maximum value

v_{\text{max}} = \omega A

. Substituting

\omega = \frac{2\pi}{0.20\pi} = 10\text{ rad/s}

into the maximum speed formula gives

v_{\text{max}} = 10 \times 0.05 = 0.50\text{ m/s}

. Computing kinetic energy yields

E_k = \frac{1}{2} (0.20) (0.50)^2 = 0.025\text{ J}

.

Adım Adım Çözüm

1

Calculate angular frequency (\omega)

\omega = 10\text{ rad/s}

Using the relation

\omega = \frac{2\pi}{T}

with

T = 0.20\pi\text{ s}

.

2

Calculate maximum speed (v_{max})

v_{max} = 0.50\text{ m/s}

Maximum velocity occurs at the equilibrium position and is given by

v_{\text{max}} = \omega A

.

3

Calculate maximum kinetic energy (E_k)

E_k = 0.025\text{ J}

Using the kinetic energy formula

E_k = \frac{1}{2} m v^2

at maximum velocity.

Anahtar Kavram

Energy conservation and maximum speed in Simple Harmonic Motion

Soru 198Soru →

A crate of mass $5\text{ kg}$ is pulled from rest along a smooth inclined plane tilted at $30^\circ$ to the horizontal by a constant force of $40\text{ N}$ acting parallel to the slope. What is the kinetic energy of the crate in Joules after moving a distance of $6\text{ m}$ along the incline? (Take $g = 10\text{ m s}^{-2}$ )

Cevabı ve açıklamayı göster

Cevap: 90

Cevap

The final kinetic energy of the crate is

90\text{ J}

.

The kinetic energy gained equals the net work done on the crate. The total work put in by the pulling force is

40 \times 6 = 240\text{ J}

, while the gravitational potential energy gained is

5 \times 10 \times (6 \sin 30^\circ) = 150\text{ J}

. Subtracting the potential energy gained from the total work gives a net kinetic energy of

240 - 150 = 90\text{ J}

.

Adım Adım Çözüm

1

Calculate the total work done by the applied force parallel to the incline

W_{\text{applied}} = 40\text{ N} \times 6\text{ m} = 240\text{ J}

Work done by a force is equal to force multiplied by displacement in the direction of the force.

2

Calculate the work done against gravity (potential energy gained)

\Delta PE = m g d \sin(30^\circ) = 5 \times 10 \times 6 \times 0.5 = 150\text{ J}

The height gained along an incline of length

d

and angle

\theta

is

h = d \sin\theta

.

3

Determine the net work done to find the final kinetic energy

KE = W_{\text{applied}} - \Delta PE = 240\text{ J} - 150\text{ J} = 90\text{ J}

According to the work-energy theorem, the net work done on an object equals its change in kinetic energy.

Anahtar Kavram

Work-Energy Theorem on an Inclined Plane

Soru 199Soru →

A body of mass $0.40\text{ kg}$ is attached to a light helical spring and set into simple harmonic motion. If the system oscillates with a period of $0.40\pi\text{ s}$ , what is the force constant of the spring in $\text{N/m}$ ?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The force constant of the spring is

10\text{ N/m}

.

The period of a mass-spring system undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{m}{k}}

. Substituting

m = 0.40\text{ kg}

and

T = 0.40\pi\text{ s}

yields

0.40\pi = 2\pi \sqrt{\frac{0.40}{k}}

. Dividing both sides by

2\pi

gives

0.20 = \sqrt{\frac{0.40}{k}}

. Squaring both sides yields

0.04 = \frac{0.40}{k}

, which gives

k = 10\text{ N/m}

.

Adım Adım Çözüm

1

Identify the period relationship for a spring-mass SHM system

T = 2\pi \sqrt{\frac{m}{k}}

This formula relates the period of oscillation

T

to the mass

m

and spring constant

k

.

2

Substitute the given mass and period into the equation

0.40\pi = 2\pi \sqrt{\frac{0.40}{k}}

Given that

m = 0.40\text{ kg}

and

T = 0.40\pi\text{ s}

.

3

Simplify the equation by isolating the square root term

\sqrt{\frac{0.40}{k}} = 0.20

Dividing both sides of the equation by

2\pi

isolates the radical.

4

Square both sides and solve for the spring constant

k

k = 10\text{ N/m}

Squaring yields

0.04 = \frac{0.40}{k}

, which rearranges to

k = \frac{0.40}{0.04} = 10\text{ N/m}

.

Anahtar Kavram

Period of oscillation of a mass-spring system in Simple Harmonic Motion

Tahmini Süre:1m 30s

Soru 200Soru →

A satellite moves in a circular orbit of radius $2R$ around a spherical planet of radius $R$ and mass $M$ . It is subsequently shifted to a larger circular orbit of radius $8R$ . What is the ratio of its initial orbital speed to its new orbital speed?

4 : 1

2 : 1

1 : 2

1 : 4

Cevabı ve açıklamayı göster

Cevap: $2 : 1$

Cevap

The ratio of the initial orbital speed to the new orbital speed is

2 : 1

.

Orbital velocity varies inversely with the square root of orbital radius (

v \propto \frac{1}{\sqrt{r}}

). Moving from radius

2R

to

8R

decreases the speed by a factor of

\sqrt{8/2} = 2

. Therefore, the ratio of initial speed to new speed is

2 : 1

.

Adım Adım Çözüm

1

Write the formula for orbital velocity

v = \sqrt{\frac{GM}{r}}

, where

G

is the gravitational constant,

M

is the mass of the planet, and

r

is the orbital radius.

Orbital speed is determined by equating gravitational force to centripetal force.

2

Set up expressions for initial and final orbital speeds

Initial speed

v_1 = \sqrt{\frac{GM}{2R}}

and final speed

v_2 = \sqrt{\frac{GM}{8R}}

.

Substitute the given orbital radii

r_1 = 2R

and

r_2 = 8R

into the formula.

3

Calculate the ratio

v_1 / v_2

\frac{v_1}{v_2} = \frac{\sqrt{\frac{GM}{2R}}}{\sqrt{\frac{GM}{8R}}} = \sqrt{\frac{8R}{2R}} = \sqrt{4} = 2

Simplifying the square root fraction gives the exact ratio.

Anahtar Kavram

Orbital Velocity and Inverse Square Law Relations

Tahmini Süre:1m 15s

Mechanics

Cevap

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