2. 14 points SerCP10 5.P.004 My Notes Ask Your Teacher A shopper in a supermarke

Question

2. 14 points SerCP10 5.P.004 My Notes Ask Your Teacher A shopper in a supermarket pushes a cart with a force of 37 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 64.0-m length aisle. (b) What is the net work done on the cart? Why? This answer has not been graded yet. (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

given F = 37 N

theta = 25 deg ( below the horizontal)

a. work dine by shopper, if d = 64 m

W = Fcos(theta)*d = 37*cos(25)*64 = 2146.13683 J

b. net work done on the cart = Wn

now, since the given force is just enough to get over the frictional force

frictional force = f

f = Fcos(theta) = 33.53338 N

work done by friction = f*d*cos(180) = -2146.13683 J

hence net work done = 0 J

c. as the speed is the same, and frictional force is the same

hence work done by friction is the same

but change in KE of the cart is 0, hence work done by the force is the same

hence

applie dforce does not change

and work dfone on the cart by the shopper remains the same as well

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