A machine which is 70 % efficient raises a 10 kg body though a certain distance and spends 100 J 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 though a certain distance and spends 100 J energy. The body is then released. On reaching the ground, the kinetic energy of the body will be (A) 0 J (B) 70 J (C) 50 J (D) 35 J

Tardigrade Admin · Accepted Answer

Energy spent by the machine to raise the body of mass

10 k g

at certain height, is

E = 100 J

Since, machine is only

70%

efficient.
Hence, energy used by machine,

E^{'} = 70%

of

E

= \frac{70}{100} \times 100 = 70 J

When body is released, then whole energy spent by machine is converted into kinetic energy on reaching the ground.

∴

Kinetic energy,

K = E^{'} = 70 J

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

0 J

70 J

50 J

35 J

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