Question

A shopper in a supermarket pushes a cart with a force of

35.0 N directed at an angle of 25.0°downward from the

horizontal. Find the work done by the shopper on the cart

as he moves down an aisle 50.0 m long

Answer 1

The 35N force applied by the shopper makes a 25 degree angle with the displacement of the cart (horizontal). The work done on the cart by the shopper is then

$W_{shoppers}=( Fcos\theta)\Delta x$

$=(35.0N)\times (50.0m)cos(25\degree)$

$=1.59\times10^{3} J$

The force exerted by the shopper is now completely horizontal and will be equal to the friction force, since the cart stays at a constant velocity, the shopper’s force had a downward vertical component, increasing the normal force on the cart, and thereby the friction force. Because there is no vertical