A shopper in a supermarket pushes a cart with a force of 35 N directed at an ang

Question

A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25° below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed.
(a) Find the work done by the shopper as she moves down a 46.0-m length aisle.
J

(b) What is the net work done on the cart?
J

Why?


(c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn't change, would the shopper's applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?

Explanation / Answer

SOLUTION: (a) Given that, Shopper in a supermarket pushes a cart with a force of , F=35 N This force directed at an angle below the horizontal is,=25o she moves down a length aisle, s=46.0 m
The work done by the shopper as she moves down a 46.0 m length aisle is        W = F S cos              =(35 N)(46 m)cos25o              =1459.15 J ________________________________________________________________ ________________________________________________________________ (b) The net work done on the cart is, W +fS Where f s= Frictional force               = -F cos Since it moves with constant speed, net force on cart is zero. Applied force in horizontal direction = - frictional force                          F cos = -f            Net work done   = W +fS                                      = 1459.15 J +(-1459.15 J)                                       = 0 J
________________________________________________________________ ________________________________________________________________ (c) The shopper's force along the horizontalcomponent would be larger since all of the force is now directed horizontally, even though the shopper is pushingwith the same force as before.
The work done would be larger (a) Given that, Shopper in a supermarket pushes a cart with a force of , F=35 N This force directed at an angle below the horizontal is,=25o she moves down a length aisle, s=46.0 m
The work done by the shopper as she moves down a 46.0 m length aisle is        W = F S cos              =(35 N)(46 m)cos25o              =1459.15 J ________________________________________________________________ ________________________________________________________________ (b) The net work done on the cart is, W +fS Where f s= Frictional force               = -F cos Since it moves with constant speed, net force on cart is zero. Applied force in horizontal direction = - frictional force                          F cos = -f            Net work done   = W +fS                                      = 1459.15 J +(-1459.15 J)                                       = 0 J
________________________________________________________________ ________________________________________________________________ (c) The shopper's force along the horizontalcomponent would be larger since all of the force is now directed horizontally, even though the shopper is pushingwith the same force as before.
The work done would be larger she moves down a length aisle, s=46.0 m
The work done by the shopper as she moves down a 46.0 m length aisle is        W = F S cos              =(35 N)(46 m)cos25o              =1459.15 J ________________________________________________________________ ________________________________________________________________ (b) The net work done on the cart is, W +fS Where f s= Frictional force               = -F cos Since it moves with constant speed, net force on cart is zero. Applied force in horizontal direction = - frictional force                          F cos = -f            Net work done   = W +fS                                      = 1459.15 J +(-1459.15 J)                                       = 0 J
________________________________________________________________ ________________________________________________________________ (c) The shopper's force along the horizontalcomponent would be larger since all of the force is now directed horizontally, even though the shopper is pushingwith the same force as before.
The work done would be larger

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