An inclined plane makes an angle of 30° with the horizontal as shown in the figure. A box of mass 20 kg is taken from point A to point B along the inclined plane of length 2 m.
(a) Calculate the potential energy gained by the box.
(b) If 10 J of work is done against friction, in moving the box from A to B then calculate the force F needed to pull the block from A to B. [g = 10 m s-2]
Show SolutionHide Solution↓
Given,
- Mass of the box (m) = 20 kg
- Incline Length (AB) = 2 m
- Incline Angle = 30°
- Work done by friction (Wf) = 10 J
- Applied force = F
- Acceleration due to gravity (g) = 10 m s-2
(a) In right angled △ ACB,
sin 30° = $\frac {\text {BC}}{\text {AB}}$
⇒ BC = AB x sin 30° = 2 x 0.5 = 1 m
So,
Potential energy gained by the box (U) = weight of the box x height gained = mg x BC = 20 x 10 x 1 = 200 J
So, potential energy gained by the box is 200 J.
(b) Let, work done by applied force be 'W'.
Then,
W = F x AB = 2 x F
Now,
W = U + Wf
= 200 + 10 = 210 J
⇒ 2 x F = 210
⇒ F = $\frac{210}{2}$
= 105 N
So, the force needed to pull the block from A to B is 105 N.