Question

A machine has a velocity ratio of 4. It is used to raise a load of 200 N by 4 m using an effort 20 kgf. Find the work done on the machine. Take g = 10 m/s${}^2$ .

• A. 1600 J
• B. 3200 J
• C. 6400 J
• D. 2000 J

Answer 1

The correct option is B

3200 J

Given, velocity ratio = 4
Load = 200 N
Effort = 20 kgf = 20 x g N = 20 x 10 N = 200 N.

Work obtained from machine (Output work) = Load $\times$ displacement = 200 $\times$ 4 = 800 J

Mechanical Advantage (MA) = $\frac{\text{load}}{\text{effort}}=\frac{200}{200}$ = 1

Efficiency = $\frac{\text{MA}}{\text{VR}}=\frac{1}{4}$

Also, efficiency = $\frac{\text{Output work}}{\text{Input work}}$

Work done on the machine (Input work) $=\frac{\text{Output work}}{\text{efficiency}}=\frac{800}{\left(1/4\right)}=3200J$