A body of mass 5 kg is moving with a velocity of 10 m s^-1. The ratio of its initial kinetic energy and final kinetic energy, if the mass of the body is doubled and its velocity is halved will be

Q. A body of mass

2 k g

is thrown upward with initial velocity

20 m / s

. After

2 s

find its kinetic energy will be:

(g = 10 m / s^{2})

Q. A constant retarding force of 50N is applied to a body of mass 10 kg moving initially with a speed of

10 m s^{- 1}

. The body comes to rest after

Join BYJU'S Learning Program

A body of mass m=10−2 kg is moving in a medium and experiences a frictional force F = −kv2. Its initial speed is v0=10ms−1. If after 10 s, its energy is 18 mv20, the value of k will be

The correct option is B 10−4 kgm−1 From the given energy, find final velocity at t=10 s. Now, use the second law of motion F=mdvdt=−kv2 and put limits to find the value of k which comes out to be 10−4 kgm−1.

A body of mass $m = 10^{- 2}$ kg is moving in a medium and experiences a frictional force F = $- k v^{2}$ . Its initial speed is $v_{0} = 10 m s^{- 1}$ . If after 10 s, its energy is $\frac{1}{8}$ $m v_{0}^{2}$ , the value of k will be

The correct option is B
$10^{- 4} k g m^{- 1}$

From the given energy, find final velocity at $t = 10 s$ .
Now, use the second law of motion $F = m \frac{d v}{d t} = - k v^{2}$ and put limits to find the value of $k$ which comes out to be $10^{- 4} k g m^{- 1}$ .