1, + 5 points osuniPhys1 7.1.P029. My Notes Ask Your 0 A shopper pushes a grocer

Question

1, + 5 points osuniPhys1 7.1.P029. My Notes Ask Your 0 A shopper pushes a grocery cart 25.0 m at constant speed on level ground, against a 37.0 N frictional force. He pushes in a direction 26.0° below the horizontal. (a) What is the work (in J) done on the cart by friction? (b) What is the work (in J) done on the cart by the gravitational force? (c) What is the work (in 3) done on the cart by the shopper? (d) Find the force the shopper exerts (in N), using energy considerations. (Enter the magnitude.) (e) What is the total work (in J) done on the cart? Additional Materials eBook

Explanation / Answer

work done by friction is always opposite in the direction of motion

wf = F*d*costheta

theta = 180 ; cos180 = -1

F = 37 N

d = 25 m

wf = -925 J

part b )

gravity works on cart perpendicularly

theta = 90 ; cos90 = 0

wg = 0 J

part c )

If the cart moves at a constant speed, no energy is transferred to it;

by work energy theorem

wf + ws = 0

ws = 925 J

part d )

ws = Fdcostheta

F = ws/dcostheta

theta = 26 degree

ws = 925 J

d = 25 m

F = 41.17 N

part e )

Since there is no change in speed, the work energy theorem says that there is no net work done on the cart

w = ws + wf = 0 J

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