Question

A person pushes a 15.5-kg shopping cart at a constant velocity for a distance of 28.5 mon a flat horizontal surface. She pushes in a direction 32.2 below the horizontal. A 30.4-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

Answer 1

a) The magnitude of the force that the shopper exerts is equal to the force of friction plus the force needed to overcome gravity. This can be calculated using the following formula:

F = ma = (15.5 kg)(9.8 m/s^2)sin(32.2) + 30.4 N

b) The work done by the pushing force is equal to the force multiplied by the distance moved in the direction of the force. This can be calculated using the following formula:

W = Fdcos(theta) = (F)(28.5 m)cos(32.2)

c) The work done by the frictional force is equal to the force of friction multiplied by the distance moved in the direction of the force. This can be calculated using the following formula:

W = Fdcos(theta) = (30.4 N)(28.5 m)cos(32.2)

d) The work done by the gravitational force is equal to the force of gravity multiplied by the distance moved in the direction of the force. This can be calculated using the following formula:

W = Fdcos(theta) = (15.5 kg)(9.8 m/s^2)sin(32.2)(28.5 m)cos(32.2)

Note that work done by force is negative when force opposes the motion.