Simple Machines Practice Questions — JAMB UTME

Question 1Question →

A screw jack with a thread pitch of $0.5\text{ cm}$ is operated using a Tommy bar of length $35\text{ cm}$ . If the machine has an efficiency of $40\%$ , what is the minimum effort force required to raise a load of mass $880\text{ kg}$ ? (Take $g = 10\text{ m/s}^2$ and $\pi = \frac{22}{7}$ )

50\text{ N}

8\text{ N}

20\text{ N}

100\text{ N}

Show answer & explanation

Answer: $50\text{ N}$

Answer

The minimum effort force required is

50\text{ N}

.

The effort distance per turn is the circumference

2\pi r = 2 \times \frac{22}{7} \times 35\text{ cm} = 220\text{ cm}

. Dividing by the pitch (

0.5\text{ cm}

) yields a Velocity Ratio of

440

. Applying the

40\%

efficiency gives a Mechanical Advantage of

\text{MA} = 0.40 \times 440 = 176

. Finally, dividing the load force (

8800\text{ N}

) by

\text{MA}

gives an effort force of

50\text{ N}

.

Step-by-Step Solution

1

Calculate the total load force in Newtons

L = m \cdot g = 880\text{ kg} \times 10\text{ m/s}^2 = 8800\text{ N}

Effort overcomes weight force, which is mass multiplied by gravitational acceleration.

2

Calculate the Velocity Ratio (VR) of the screw jack

\text{VR} = \frac{2 \pi r}{p} = \frac{2 \times \frac{22}{7} \times 35\text{ cm}}{0.5\text{ cm}} = \frac{220\text{ cm}}{0.5\text{ cm}} = 440

The effort travels around the circumference of a circle of radius

r

, while the load moves vertically by one pitch length

p

per revolution.

3

Determine the Mechanical Advantage (MA) from efficiency

\text{MA} = \eta \times \text{VR} = 0.40 \times 440 = 176

Efficiency is defined as

\eta = \frac{\text{MA}}{\text{VR}}

, so

\text{MA} = \eta \cdot \text{VR}

.

4

Calculate the effort force E

E = \frac{L}{\text{MA}} = \frac{8800\text{ N}}{176} = 50\text{ N}

Mechanical advantage is the ratio of load to effort force

(\text{MA} = \frac{L}{E})

.

Key Concept

Relationship between Velocity Ratio, Mechanical Advantage, Efficiency, and Pitch in a Screw Jack

Question 2Question →

An inclined plane of length $5\text{ m}$ is used to lift a load of $400\text{ N}$ through a vertical height of $1\text{ m}$ . If an effort of $100\text{ N}$ is applied parallel to the incline, what is the efficiency of the machine?

80\%

125\%

25\%

20\%

Show answer & explanation

Answer: $80\%$

Answer

The efficiency of the machine is

80\%

.

The correct answer of

80\%

is obtained by finding the velocity ratio (distance moved by effort divided by distance moved by load,

5 / 1 = 5

) and the mechanical advantage (load divided by effort,

400 / 100 = 4

), then calculating efficiency as

(\text{MA} / \text{VR}) \times 100\% = (4 / 5) \times 100\% = 80\%

.

Step-by-Step Solution

1

Calculate the Velocity Ratio (VR) of the inclined plane

\text{VR} = \frac{\text{Length of incline}}{\text{Height}} = \frac{5\text{ m}}{1\text{ m}} = 5

Velocity Ratio is the distance moved by the effort divided by the distance moved by the load.

2

Calculate the Mechanical Advantage (MA)

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{400\text{ N}}{100\text{ N}} = 4

Mechanical Advantage measures how many times a machine multiplies the applied force.

3

Calculate the Efficiency

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

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

Key Concept

Efficiency of Simple Machines (Inclined Plane)

Estimated Time:45s

Question 3Question →

A wheel and axle machine having a wheel radius of $25\text{ cm}$ and an axle radius of $5\text{ cm}$ is used to raise a load of mass $80\text{ kg}$ vertically through a height of $10\text{ m}$ . If the efficiency of the machine is $80\%$ , what is the work done against friction during this operation? (Take $g = 10\text{ m s}^{-2}$ )

1,600\text{ J}

2,000\text{ J}

8,000\text{ J}

10,000\text{ J}

Show answer & explanation

Answer: $2,000\text{ J}$

Answer

2,000\text{ J}

The useful work done on the load is

W_{\text{out}} = mgh = 80 \times 10 \times 10 = 8000\text{ J}

. Since efficiency is

80\%

, the total work input required is

W_{\text{in}} = \frac{8000}{0.80} = 10,000\text{ J}

. Therefore, the work lost to friction is

W_{\text{in}} - W_{\text{out}} = 10,000\text{ J} - 8,000\text{ J} = 2,000\text{ J}

.

Step-by-Step Solution

1

Calculate the useful work output of the machine

W_{\text{out}} = mgh = 80\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 8,000\text{ J}

Useful work output is the gravitational potential energy gained by raising the load.

2

Calculate the total work input using the efficiency equation

W_{\text{in}} = \frac{W_{\text{out}}}{\text{Efficiency}} = \frac{8,000\text{ J}}{0.80} = 10,000\text{ J}

Efficiency is defined as

\frac{\text{Work Output}}{\text{Work Input}}

, so Work Input =

\frac{\text{Work Output}}{\text{Efficiency}}

.

3

Determine the work done against friction

W_{\text{friction}} = W_{\text{in}} - W_{\text{out}} = 10,000\text{ J} - 8,000\text{ J} = 2,000\text{ J}

The energy wasted as heat and friction is the difference between total work input and useful work output.

Key Concept

Efficiency of Simple Machines and Work Done Against Friction

Question 4Question →

A hydraulic press consists of a small effort piston of radius $2\text{ cm}$ and a large load piston of radius $8\text{ cm}$ . When an effort force of $50\text{ N}$ is applied to the small piston, it successfully lifts a load of $600\text{ N}$ placed on the large piston. What is the efficiency of this hydraulic press?

75\%

133\%

33\%

25\%

Show answer & explanation

Answer: $75\%$

Answer

The efficiency of the hydraulic press is

75\%

.

The correct answer is

75\%

. The mechanical advantage is

MA = \frac{600}{50} = 12

, and the velocity ratio is

VR = \left(\frac{8}{2}\right)^2 = 16

. Taking the ratio

\frac{MA}{VR} \times 100\%

yields

\frac{12}{16} \times 100\% = 75\%

.

Step-by-Step Solution

1

Calculate the Mechanical Advantage (MA) of the hydraulic press.

MA = \frac{\text{Load}}{\text{Effort}} = \frac{600\text{ N}}{50\text{ N}} = 12

Mechanical advantage measures the force multiplication factor of a machine.

2

Calculate the Velocity Ratio (VR) of the hydraulic press using the piston radii.

VR = \frac{\text{Area of load piston}}{\text{Area of effort piston}} = \frac{\pi R^2}{\pi r^2} = \left(\frac{8\text{ cm}}{2\text{ cm}}\right)^2 = 4^2 = 16

For a hydraulic press, velocity ratio equals the ratio of the cross-sectional areas of the pistons, which simplifies to the square of the radii ratio.

3

Compute the efficiency of the machine.

\eta = \frac{MA}{VR} \times 100\% = \frac{12}{16} \times 100\% = 75\%

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

Key Concept

Mechanical Advantage, Velocity Ratio, and Efficiency of a Hydraulic Press

Estimated Time:1m 30s

Question 5Question →

A block and tackle system consisting of $5$ pulleys is used to raise a load of $200\text{ N}$ through a vertical height of $4\text{ m}$ . If the efficiency of the system is $80\%$ , what is the work done against friction, in joules, during the lifting process?

Show answer & explanation

Answer: 200

Answer

The work done against friction during the lifting process is

200\text{ J}

.

The useful work output is achieved by lifting the

200\text{ N}

load through a vertical height of

4\text{ m}

, yielding

W_{\text{out}} = 200 \times 4 = 800\text{ J}

. Given an efficiency of

80\%

(

0.80

), the total work input required from the effort force is

W_{\text{in}} = \frac{800}{0.80} = 1000\text{ J}

. The energy lost to overcome frictional resistance in the pulleys equals the total work input minus useful work output:

1000\text{ J} - 800\text{ J} = 200\text{ J}

.

Step-by-Step Solution

1

Calculate useful work output

W_{\text{out}} = 800\text{ J}

Useful work output is the energy required to raise the load through the specified height (

W_{\text{out}} = \text{Load} \times \text{height}

).

2

Determine total work input

W_{\text{in}} = 1000\text{ J}

The total work input is calculated from the efficiency formula:

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

.

3

Compute work done against friction

W_{\text{friction}} = 200\text{ J}

The work lost overcoming friction is the difference between total work input and useful work output (

W_{\text{in}} - W_{\text{out}}

).

Key Concept

Work and Efficiency in Pulley Systems

Estimated Time:1m 30s

Question 6Question →

A machine with a velocity ratio of $6$ is used to raise a load of $540\text{ N}$ by applying an effort of $120\text{ N}$ . What is the efficiency of the machine?

75\%

133.3\%

22.2\%

450\%

Show answer & explanation

Answer: $75\%$

Answer

The efficiency of the machine is

75\%

.

The correct option is

75\%

. Mechanical Advantage is found by dividing the load (

540\text{ N}

) by the effort (

120\text{ N}

), giving

4.5

. Dividing this mechanical advantage by the given velocity ratio (

6

) and expressing as a percentage yields

\frac{4.5}{6} \times 100\% = 75\%

.

Step-by-Step Solution

1

Calculate Mechanical Advantage (MA)

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{540\text{ N}}{120\text{ N}} = 4.5

Mechanical advantage measures the force magnification produced by the machine.

2

Calculate Efficiency (\eta)

\eta = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100\% = \left(\frac{4.5}{6}\right) \times 100\% = 75\%

Efficiency is the ratio of Mechanical Advantage to Velocity Ratio multiplied by 100%.

Key Concept

Efficiency of a Simple Machine

Estimated Time:1m 0s

Question 7Question →

An inclined plane is set at an angle of $30^\circ$ to the horizontal. An effort of $320\text{ N}$ applied parallel to the plane is used to push a load of $480\text{ N}$ up the incline at a constant speed. What is the efficiency of the inclined plane expressed as a percentage?

Show answer & explanation

Answer: 75%; 75; 75 percent

Answer

The efficiency of the inclined plane is

75\%

.

For an inclined plane inclined at

30^\circ

to the horizontal, the velocity ratio (VR) is given by

\text{VR} = \frac{1}{\sin 30^\circ} = 2

. The mechanical advantage (MA) is

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{480\text{ N}}{320\text{ N}} = 1.5

. Dividing MA by VR and multiplying by

100\%

yields an efficiency of

\left(\frac{1.5}{2}\right) \times 100\% = 75\%

.

Step-by-Step Solution

1

Determine the Velocity Ratio (VR) of the inclined plane from its angle of inclination

VR = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2

For an inclined plane with inclination angle \theta, the velocity ratio is equal to \frac{1}{\sin \theta}.

2

Calculate the Mechanical Advantage (MA)

MA = \frac{\text{Load}}{\text{Effort}} = \frac{480\text{ N}}{320\text{ N}} = 1.5

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

3

Calculate the efficiency of the machine

\text{Efficiency} = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{1.5}{2} \times 100\% = 75\%

Efficiency is the ratio of mechanical advantage to velocity ratio expressed as a percentage.

Key Concept

Efficiency of an Inclined Plane

Question 8Question →

A screw jack with a pitch of $4\text{ mm}$ and a tommy bar of length $56\text{ cm}$ is used to raise a heavy load of mass $1100\text{ kg}$ . If an effort force of $50\text{ N}$ is applied at the outer end of the tommy bar, what is the efficiency of the screw jack? (Take $\pi = \frac{22}{7}$ and $g = 10\text{ m/s}^2$ )

25%

400%

157%

2.5%

Show answer & explanation

Answer: 25%

Answer

The efficiency of the screw jack is 25%

The mechanical advantage is calculated as

MA = \frac{11,000\text{ N}}{50\text{ N}} = 220

. The velocity ratio for a screw jack is

VR = \frac{2 \pi R}{p} = \frac{2 \times \frac{22}{7} \times 0.56\text{ m}}{0.004\text{ m}} = 880

. Dividing

MA

by

VR

and multiplying by

100\%

yields an efficiency of

\frac{220}{880} \times 100\% = 25\%

.

Step-by-Step Solution

1

Calculate the load force (weight) from the given mass.

W = m \times g = 1100\text{ kg} \times 10\text{ m/s}^2 = 11,000\text{ N}

Mass must be converted to weight force in newtons to compute mechanical advantage.

2

Calculate the Mechanical Advantage (MA).

MA = \frac{\text{Load}}{\text{Effort}} = \frac{11,000\text{ N}}{50\text{ N}} = 220

Mechanical advantage is the ratio of output load force to input effort force.

3

Calculate the Velocity Ratio (VR) of the screw jack.

VR = \frac{2 \pi R}{p} = \frac{2 \times \frac{22}{7} \times 0.56\text{ m}}{0.004\text{ m}} = \frac{3.52\text{ m}}{0.004\text{ m}} = 880

Velocity ratio is the distance moved by the effort in one full revolution (

2\pi R

) divided by the distance moved by the load in one revolution (pitch

p

).

4

Calculate the efficiency of the machine.

\text{Efficiency} = \frac{MA}{VR} \times 100\% = \frac{220}{880} \times 100\% = 25\%

Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio multiplied by 100%.

Key Concept

Mechanical Advantage, Velocity Ratio, and Efficiency of a Screw Jack

Question 9Question →

A wheel and axle system consists of a wheel with a radius of $25\text{ cm}$ attached to an axle with a radius of $5\text{ cm}$ . What is the velocity ratio of this machine?

Show answer & explanation

Answer: 5

Answer

The velocity ratio of the machine is 5.

The velocity ratio (VR) of a wheel and axle is the ratio of the radius of the wheel (

R

) to the radius of the axle (

r

). Calculating

\frac{25\text{ cm}}{5\text{ cm}}

gives a velocity ratio of

5

.

Step-by-Step Solution

1

Identify the formula for the velocity ratio of a wheel and axle system.

\text{VR} = \frac{R}{r}

Velocity ratio is defined as the distance moved by the effort (proportional to wheel radius) divided by the distance moved by the load (proportional to axle radius).

2

Substitute the given values into the formula.

\text{VR} = \frac{25\text{ cm}}{5\text{ cm}}

The radius of the wheel

R = 25\text{ cm}

and the radius of the axle

r = 5\text{ cm}

.

3

Calculate the final ratio.

\text{VR} = 5

Dividing 25 by 5 yields 5. The ratio is dimensionless because the units of centimeters cancel out.

Key Concept

Velocity Ratio of a Wheel and Axle

Question 10Question →

A wheel and axle machine has a wheel of radius $25\text{ cm}$ and an axle of radius $5\text{ cm}$ . If an effort force of $100\text{ N}$ is applied to lift a load of $400\text{ N}$ , what is the efficiency of the machine?

80\%

125\%

20\%

5\%

Show answer & explanation

Answer: $80\%$

Answer

The efficiency of the wheel and axle machine is

80\%

.

First, calculate the velocity ratio as the ratio of wheel radius to axle radius:

\text{VR} = \frac{25}{5} = 5

. Next, calculate mechanical advantage as load over effort:

\text{MA} = \frac{400}{100} = 4

. Finally, calculate efficiency by dividing mechanical advantage by velocity ratio:

\text{Efficiency} = \frac{4}{5} \times 100\% = 80\%

.

Step-by-Step Solution

1

Calculate the Velocity Ratio (VR) of the wheel and axle system

\text{VR} = \frac{\text{Radius of Wheel}}{\text{Radius of Axle}} = \frac{25\text{ cm}}{5\text{ cm}} = 5

For a wheel and axle, the velocity ratio is the ratio of the radius of the wheel to the radius of the axle.

2

Calculate the Mechanical Advantage (MA) of the machine

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{400\text{ N}}{100\text{ N}} = 4

Mechanical advantage measures the force magnification of a machine, defined as the ratio of load raised to effort applied.

3

Calculate the Efficiency of the machine

\text{Efficiency} = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{4}{5} \times 100\% = 80\%

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

Key Concept

Efficiency of Simple Machines

Question 11Question →

A hydraulic press has a small piston with a diameter of $4\text{ cm}$ and a large piston with a diameter of $20\text{ cm}$ . If an effort force of $80\text{ N}$ applied to the small piston raises a load of $1500\text{ N}$ placed on the large piston, what is the efficiency of the machine?

133.3\%

75.0\%

375.0\%

26.7\%

Show answer & explanation

Answer: $75.0\%$

Answer

The efficiency of the hydraulic press is

75.0\%

.

The mechanical advantage is

MA = \frac{1500}{80} = 18.75

. The velocity ratio for pistons of diameters

20\text{ cm}

and

4\text{ cm}

is

VR = \left(\frac{20}{4}\right)^2 = 25

. Efficiency is

\frac{MA}{VR} \times 100\% = \frac{18.75}{25} \times 100\% = 75.0\%

.

Step-by-Step Solution

1

Calculate Mechanical Advantage (MA)

MA = \frac{\text{Load}}{\text{Effort}} = \frac{1500\text{ N}}{80\text{ N}} = 18.75

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

2

Calculate Velocity Ratio (VR) for the hydraulic press

VR = \frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{20\text{ cm}}{4\text{ cm}}\right)^2 = 5^2 = 25

The velocity ratio of a hydraulic press equals the ratio of the cross-sectional areas of the pistons, which simplifies to the square of the ratio of their diameters.

3

Calculate Efficiency (\eta)

\eta = \frac{MA}{VR} \times 100\% = \frac{18.75}{25} \times 100\% = 75.0\%

Efficiency is the ratio of mechanical advantage to velocity ratio expressed as a percentage.

Key Concept

Hydraulic Press Efficiency and Velocity Ratio

Question 12Question →

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?

Show answer & explanation

Answer: 4; 4.0

Answer

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

.

Step-by-Step Solution

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.

Key Concept

Mechanical Advantage of Simple Machines

Estimated Time:45s

Question 13Question →

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?

Show answer & explanation

Answer: 80%; 80; 80 percent; 80 %

Answer

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\%

.

Step-by-Step Solution

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.

Key Concept

Efficiency of a Simple Machine

Estimated Time:1m 30s

Question 14Question →

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?

Show answer & explanation

Answer: 75%; 75 %; 75 percent; 75

Answer

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%.

Step-by-Step Solution

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\%

.

Key Concept

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

Estimated Time:1m 30s

Question 15Question →

An inclined plane of length $10\text{ m}$ is used to raise a heavy crate of load $900\text{ N}$ to a height of $2\text{ m}$ . If an effort force of $250\text{ N}$ is applied parallel to the inclined surface to push the crate up at constant speed, what is the efficiency of the simple machine?

$72\%

$139\%

$36\%

$5.6\%

Show answer & explanation

Answer: $72\%

Answer

The efficiency of the inclined plane is

72\%

.

The mechanical advantage of the machine is

MA = \frac{900\text{ N}}{250\text{ N}} = 3.6

, and its velocity ratio is

VR = \frac{10\text{ m}}{2\text{ m}} = 5

. Dividing

MA

by

VR

and multiplying by

100\%

gives an efficiency of

72\%

.

Step-by-Step Solution

1

Calculate the Mechanical Advantage (MA)

MA = \frac{\text{Load}}{\text{Effort}} = \frac{900\text{ N}}{250\text{ N}} = 3.6

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

2

Calculate the Velocity Ratio (VR) of the inclined plane

VR = \frac{\text{Length of plane}}{\text{Height of plane}} = \frac{10\text{ m}}{2\text{ m}} = 5

Velocity ratio for an inclined plane is the ratio of distance moved by effort along the incline to distance moved by load vertically.

3

Calculate the Efficiency (\eta)

\eta = \left(\frac{MA}{VR}\right) \times 100\% = \left(\frac{3.6}{5}\right) \times 100\% = 72\%

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

Key Concept

Efficiency of an Inclined Plane

Simple Machines

Answer

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