A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 7.2 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall

480N

Explanation:

Given data

M1= 1.2kg

U1= 8m/s

V1=7.2 m/s

T= 2ms= 0.002

We know that the impulse expression is

P=FT

M(U1-V1)= FT

substitute

1.2(8-7.2)= F*0.002

1.2*0.8=  F*0.002

0.96=  F*0.002

F= 0.96/0.002

F=480N

Hence the force is 480N

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one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.