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antiseptic1488 [7]
3 years ago
7

A 70 kg box undergoes a horizontal acceleration of 3.0 m/s^2 on a level surface when pulled by a 130 N force, What is the coeffi

cient of kinetic friction between the box and the surface?
Physics
1 answer:
elixir [45]3 years ago
7 0

I think there's a typo in the question...

Vertically, the box is in equilibrium, so its weight and the normal force (magnitudes <em>w</em> and <em>n</em>, respectively) are such that

<em>n</em> + (-<em>w</em>) = 0

The box has mass 70 kg, and assuming gravitational acceleration with magnitude <em>g</em> = 9.80 m/s², it has a weight of

<em>w</em> = (70 kg) <em>g</em> = 686 N

and hence

<em>n</em> = 686 N

Horizontally, the box is accelerated 3.0 m/s², so the net force acting on it is

∑ <em>F</em> = (70 kg) (3.0 m/s²) = 210 N

and the only forces acting in this dimension are the pulling force with magnitude 130 N and the friction force with magnitude <em>f</em> so that

130 N - <em>f</em> = 210 N

The friction force is proportional to the normal force by a factor of <em>µ</em>, the coefficient of kinetic friction, so that

<em>f</em> = <em>µ</em> <em>n</em>

and so we have

130 N - <em>µ</em> (686 N) = 210 N   →   <em>µ</em> ≈ -0.117

but <em>µ</em> can't be negative!

The problem is that the pulling force should have a magnitude larger than that of the net force, so either the given mass, acceleration, or pulling force are incorrect.

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