A shopper in a supermarket pushes a cart with a force of35 N directed at an angl

Question

A shopper in a supermarket pushes a cart with a force of35 N directed at an angle of 25o below the horizontal. The force is justsufficient to overcome various frictional forces, so the cart movesat a constant speed.

(a) Find the work done by he shopper as she moves down a50.0-m length aisle.

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

(c) The shopper goes down the next aisle, pushing horizontally andmaintaining the same speed as before. If the work done byfrictional forces doesn't change, would the shopper's applied forceby larger, smaller, or the same? What about the work done on thecart 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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