A support pushes a grocery cart 21.0 m at constant speed on level ground, agains

Question

A support pushes a grocery cart 21.0 m at constant speed on level ground, against a 32.0 N frictional force. He pushes in a direction 26.0 degree 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 J) 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?

Explanation / Answer

(a)

The work done by friction is in the opposite direction of the motion, so theta = 180,

therefore,

Wf = Fd*cos(theta)

Wf = 32 x 21 x cos(180)

Wf = -672 J

(b)

The work done by gravity is perpendicular to the direction of motion, so theta = 90

Wg = 32 x 21 x cos(90)

Wg = 0 J

If the cart moves at a constant speed, no energy is transferred to it, from the work-energy theorem:

Wnet = Ws + Wf

Ws + Wf = 0

Ws = +672 J

(d)

W = Fd*cos(theta)

theta = 26

F = W / d*cos(theta)

F = 672 / (21 x cos(26))

F = 35.6 N

(e)

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

Wnet = Wf + Ws = -672 J + 672 J = 0 J

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