Question

124. If the centripetal force acting on a body revolving along a circular path of radius 25 m is 200 N, its KE is 1) 2.5 KJ 2) 4 KJ 3) 4.8KJ 4) 6J​

Answer 1

Answer:

1.) 2.5 KJ

Explanation:

given,

f = 200 N

r = 25 m

$f = \frac{mv^{2} }{r}$

$200 = \frac{mv^{2} }{25}$

$mv^{2} = 200 \times 25$

      $=5000 kgm/sec^{2} \\= 5 \times 10^{3 } kgm/sec^{2}$

$K.E. = \frac{1}{2} mv^{2} \\= \frac{1}{2} \times 5 \times 10^{3}$

$= 2.5 \times 10^{3} KJ$

So, option (1) 2.5 KJ is correct answer