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AysviL [449]
3 years ago
6

A box of mass m is pushed horizontally on a rough floor with an initial speed of 2 m/s. The coefficient of kinetic friction betw

een the surface and the box is µ_k = 0.1. Calculate the distance the box will move before stopping.
Physics
1 answer:
dusya [7]3 years ago
7 0
<span>The box is pushed initially.
F = ma;
m - mass of box
a - acceleration. 
</span><span>
ma = -mu_k mg
</span><span>a = -mu_k g
</span><span>
Equation of motion: v^2 = u^2 + 2as
</span><span>
The box comes to a rest, v = 0 m/s
</span><span>
Here, u = 2 m/s and a = -`mu_k g = - 0.1 xx 9.81 = - 0.981 m/s^2
</span><span>Thus, u^2 = -2as
</span><span>s = u^2/(-2a) = (2^2) / (-2 x -0.981) = 2.04 m. 
</span><span>
So the box will move a distance of 2.04 m over the rough surface.</span>
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