Question

When a constant force acts of 10sec on a body of mass 10kg,which is initially at rest, moves a distance of 500cm in 10sec. Calculate the frictional force required to bring the body to rest.

Answer 1

Answer:

the answer is 0.5 N.

Explanation:

time = 10 sec

mass = 10 kg

displacement = 500 cm

v = d/t = 500/10 = 50 cm/s = 0.5 m/s

force of the body = m*a

a = (v-u)/t

a = (0.5-0)/10

a = 0.5/10 = 0.05 m/s²

therefore force of friction required = 10*0.05 = 0.5 N

DO MARK IT AS BRAINLIEST.

Answer 2

Answer:

I don't know answer of this question