A clerk pushes a filing cabinet of mass 22.0 kg accross the floor by exerting a horizontal force of magnitude 98 N. The magnitude of the force of kinetic friction acting on the cabinet is 87 N. The cabinet starts from rest. Use the law of conservation of energy to determine the speed of the cabinet after it moves 1.2 m.

The net force acting is F = 98 - 87 = 11 N

The acceleration is a = F/m = 0.5 m/s^2

Conservation of energy says that
(1/2) m V^2 = F X
V = sqrt [2 (F/m) X] = sqrt (2 a X)

After moving a distance X = 1.2 m, the speed is V = sqrt (2 a X)= sqrt(2*0.5%1.2) = 1.1 m/s

Thank you so much. I just checked the back of my textbook and that is the right answer! :)

1 answer

  1. The change in kinetic energy of the cabinet is equal to the net work done on the cabinet, which is equal to the force of kinetic friction times the distance over which the force acts:

    $\Delta KE = F_{k}d = (87\text{ N})(1.2\text{ m}) = 104.4\text{ J}$

    The initial kinetic energy of the cabinet is zero since it starts from rest, so the final kinetic energy of the cabinet is also 104.4 J. The kinetic energy of the cabinet is given by

    $KE = \frac{1}{2}mv^{2}$

    where $m$ is the mass of the cabinet and $v$ is its speed. Solving for $v$, we find

    $v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2(104.4\text{ J})}{22.0\text{ kg}}} = \boxed{3.0\text{ m/s}}$

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