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]

Given,

• Mass of the box (m) = 20 kg
• Incline Length (AB) = 2 m
• Incline Angle = 30°
• Work done by friction (W_f) = 10 J
• Applied force = F
• Acceleration due to gravity (g) = 10 m s^-2

(a) In right angled △ ACB,

sin 30° = $\dfrac {\text {BC}}{\text {AB}}$

⇒ 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 + W_f

= 200 + 10 = 210 J

⇒ 2 x F = 210

⇒ F = $\dfrac{210}{2}$

= 105 N

So, the force needed to pull the block from A to B is 105 N.