4. . A shopper in a supermarket pushes a cart with a force of 35 N directed at a

Question

4. . A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25 degree 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 clown a 50.0-m length aisle. (b) What is the net work clone on the cart? 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

frictional force = F*cos(25) = 35*cos(25) = 31.72 N

using work energy principle

Workdone by the net force = change in KE

W= 0.,5*m*(v^2-u^2)

for constant speedv =u

then W = 0 J

A) work done by the shopper is F*cos(25)*S = 35*cos(25)*50 = 1586.03 J

B) work done by the frictional force is 35*cos(25)*50*cos(180) = -1586.03 J

net work done is 0 J

C) shopper's horizontal force is same as the frictional force

W_shopper = 31.72*50 = 1586 J

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