Question

The length of a smooth inclined plane of 30-degree inclination is 5 m. The work done in moving a 10 kg mass from the bottom of the inclined plane to the top is _____ Joules. (Assume g = 10m/s^2)

(a) 250

(b) 1000

(c) 1250

(d) 500

I got this question in examination.

My doubt is from Work Done by a Variable Force topic in section Work, Energy and Power of Physics – Class 11

Answer 1

The correct answer is (a) 250

Explanation: Length (l) = 5 m

Height = Displacement = 5 x sin30

 = 5 x (1/2)

 = 2.5

Force on object = m x g

= 10 X 10

 = 100 N

Work = Force x Displacement

 = 100 x 2.5

 = 250 Joules.