Question

A box of mass 50 kg is pulled up on an incline 12m long and 2m high by a constant force of 100N from rest. It acquires a velocity of 2m/s on reaching the top. Work done against friction is:(g=10m/ss)
Options:
(a)50J
(b)100J
(c)150J
(d)200J

Answer 1

The answer is option (b) 100J .

Answer 2

Answer:

(b) 100 J

Explanation:

As per the given question,

According to the law of conservation of energy:

Total work done by pulling is equal to that sum of work done due to potential energy, work done due to kinetic energy, and work done due to friction.

Therefore,

Total work  = W(K.E) + W(P.E) + W(friction)

$W = mgh + \frac{1}{2}mv^{2}+W_{f}$

$Fl= mgh + \frac{1}{2}mv^{2}+W_{f}$

$100\times 12= 50\times 10\times 2 + \frac{1}{2}50\times 2^{2}+W_{f}$

$1200=1000 +100+W_{f}$

$W_{f} = 100\ joule$

Hence, Work done against friction is = 100 Joule.