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A 5.00-kg body accelerates from 3.25 m/s to 9.75 m/s in 3.0 seconds when a force acts upon it. Find (a) the change in momentum of the body, (b) the impulse produced by the force, and (c) the magnitude of the force.

(a) the change in momentum of the body is 32.5 kg.m/s

(b) the impulse produced by the force is 32.5 N.s

(c) the magnitude of the force is 10.83 N

Explanation:

Given;

mass of the body, m =  5.00-kg

initial velocity of the body, u = 3.25 m/s

final velocity of the body, v = 9.75 m/s

time taken, t = 3.0 seconds

Part (a) the change in momentum of the body

ΔP = mv - mu

     = (5 x 9.75) - (5 x 3.25)

     = 48.75 - 16.25

     = 32.5 kg.m/s

Part (b) the impulse produced by the force

I = f x t

where;

I is impulse

f is the applied force

t is time

f x t = mΔv

I = mΔv

I = m (v -u)

I = 5 (9.75 - 3.25)

I = 32.5 N.s

Part (c) the magnitude of the force;

I = f x t

f = I / t

where;

f is magnitude of the force.

I is impulse

t is time

f = 32.5 / 3

f = 10.83 N

answer( sicar );

friction

explanation:

they us to move forward and backwards