PHYSICS

JAMB 2000 - Question 10

Physics 2000 JAMB Past Questions - Question 10: The velocity ratio of a machine is 5 and its efficiency is 75%. What effort would be needed to lift a load of 150 N with the machine?

The velocity ratio of a machine is 5 and its efficiency is 75%. What effort would be needed to lift a load of 150 N with the machine?
A:
B:
C:
D:
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Correct Answer

B

Explanation

Efficiency=L/E *1/V.R *10075=150/E *1/5 *100/1 ;E=40NTo calculate the effort needed to lift a load using a machine, you can use the formula for mechanical advantage (MA) and the formula for efficiency.To determine the effort needed to lift a load of 150 N with the machine, we need to use the concepts of velocity ratio and efficiency.

The velocity ratio (VR) of a machine is defined as the ratio of the distance moved by the effort to the distance moved by the load. In this case, the velocity ratio is given as 5.

Efficiency (η) of a machine is defined as the ratio of the useful work output to the total work input, expressed as a percentage. In this case, the efficiency is given as 75% or 0.75.

To calculate the effort (E), we can use the formula:

Efficiency = (Load × Load distance) / (Effort × Effort distance)

Given:
Load = 150 N
Velocity Ratio (VR) = 5
Efficiency (η) = 75% or 0.75

Let's assume the effort distance and load distance are equal (d) for simplicity.

Using the formula, we can rearrange it to solve for the effort (E):

Efficiency = (Load × d) / (Effort × d)

0.75 = (150 N) / (E × d)

Simplifying the equation, we get:

E × d = (150 N) / 0.75

E × d = 200 N

Since the effort distance (d) and load distance are equal, we can substitute d with 1:

E × 1 = 200 N

E = 200 N

Therefore, the effort needed to lift a load of 150 N with the machine is 200 N.

The mechanical advantage (MA) of a machine is the ratio of the load force (output force) to the effort force (input force). In this case, the velocity ratio is also known, and it's equal to the mechanical advantage. So, MA = 5.

Efficiency (Eff) is given as 75%, but it should be expressed as a decimal, so Eff = 0.75.

The formula for calculating the effort force (F_effort) is:

F_effort = (Load force) / (MA * Eff)

Given:
Load force (F_load) = 150 N
MA = 5
Eff = 0.75

Now, plug these values into the formula:

F_effort = (150 N) / (5 * 0.75)

F_effort = 150 N / 3.75

F_effort = 40 N