Question

block of 100 kg is kept on the horizontal plane of coefficient of friction 0.5 the amount of work done in dragging it with a velocity of 10 metre per second for 10 seconds

A.49×10^3j

B.0.49×10^5j

C.4.9×10^4j

D.98×1\2×10^6j

Answer 1

Answer: A

Explanation: distance= velocity × time

Thus distance=10×10=100m

W=f×s

f = coefficient of friction ×mass×g

f= 0.5×100×9.8=490N

W= 490×100

W= 49000J

Answer 2

Answer:

The amount of work done in dragging it with a velocity of 10 metre per second for 10 seconds is 49050 J

Explanation:

Given:

Velocity = v = 10m/s

time = 10 seconds

So distance = v*t = 10*10=100 m

Work done = Force*distance

Force = coefficient of friction*mass*acceleration due to gravity

         = 0.5*100*9.81 = 490.5 N

So, Workdone = 490.5*100 = 49050 J