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Question

A boy pushes a box of mass 2 kg with a force of F=(20^i+10^j) N on a frictionless surface. If the box was initially at rest, then __________m is the displacement along the xaxis after 10 s.


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Solution

Given:
F=(20^i+10^j) N
m=2 kg

So, acceleration,
a=Fm=20^i+10^j2=(10^i+5^j) m/s2

Now,
Sx=uxt+12axt2

Sx=0×10+12×10×102

Sx=500 m

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