A machine which is 70 % efficient raises a 10 kg body through a certain distance and spends 100 J of energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

Question

A machine which is 70 % efficient raises a 10 kg body through a certain distance and spends 100 J of energy. The body is then released. On reaching the ground, the kinetic energy of the body will be (A) 50 J (B) 35 J (C) 70 J (D) 00 J

Tardigrade Admin · Accepted Answer

70 % efficiency means body will get 70 % of energy spent by the machine =(70/100) × 100 J =70 J Hence P.E. of body =70 J On reaching ground PE becomes KE =70 J.

Q. A machine which is $70%$ efficient raises a $10 k g$ body through a certain distance and spends $100 J$ of energy. The body is then released. On reaching the ground, the kinetic energy of the body will be

50 J

35 J

70 J

00 J

$70%$ efficiency means body will get $70%$ of
energy spent by the machine $= \frac{70}{100} \times 100 J = 70 J$
Hence P.E. of body $= 70 J$ On reaching ground PE becomes $K E = 70 J$ .

Q. A machine which is 70% efficient raises a 10kg body through a certain distance and spends 100J of energy. The body is then released. On reaching the ground, the kinetic energy of the body will be