Question

A cart of weight $\text{10 N}$ is pulled $\text{1.2 m}$ up a ramp by an applied force of magnitude $\text{5.0 N}$ (parallel to the ramp), raising the cart $\text{0.50 m}$ above its initial level. Calculate percentage efficiency of the work?

• A. $\text{83.33\%}$
• B. $\text{85\%}$
• C. $\text{84.5\%}$
• D. $\text{86.33\%}$

Answer 1

The correct option is A

$\text{83.33\%}$

Given, weight of the cart (mg) = $\text{10 N}$, vertical displacement (h)= $\text{0.5 m}$ and distance covered on the ramp (s)=$1.2m$.

Energy output = Potential energy
$=m\times g\times h$
$=10\times 0.5=5J$

Energy input = Work done to move the cart
= Force $\times$ distance covered on the ramp
= 5 $\times$ 1.2 = 6 J
Percentage efficiency $=\frac{\text{Energy output}}{\text{Energy input}}$
= $\frac{5}{6}$ $\times 100$
$\text{= 83.33\%}$