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
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
"=(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
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