Question

A force actes for 10 seconds on the body of mass 10 kg after which the force stops thus body describes 50 m in next 5 sec find the magnitude of force

Answer 1

Answer:

20N

Explanation:

As after stopping the final velocity of the first case will become initial velocity of second case so:(case 1-A force acts for 10 seconds on the body of mass 10 kg after which the force stops    case 2-body describes 50 m in next 5 sec )

For second case:

s=ut+1/2at²

° s=ut+1/2(v-u/t)t²

50=u(5)+1/2(0-u/5)5²

50=5u+1/2(-u/5)25                

50=5u+1/2(-u)(5)          (canceling 5 and 25)

50=5u+(-5u/2)

50=5u-5u/2

50=10u/2-5u/2             (taking LCM)

50=5u/2

50*2=5u

100/5=u

u=20m/s    

now taking u as v in first case

F=ma

F=10*(20-0/10)

F=10*(20/10)

F=10*2

F=20N