Question

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 coefficient of kinetic friction between the box and the surface?

Answer 1

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

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

n + (-w) = 0

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

w = (70 kg) g = 686 N

and hence

n = 686 N

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

∑ F = (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 f so that

130 N - f = 210 N

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

f = µ n

and so we have

130 N - µ (686 N) = 210 N   →   µ ≈ -0.117

but µ 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.