A shopper pushes a grocery cart 17.0 m at constant speed on level ground, agains

Question

A shopper pushes a grocery cart 17.0 m at constant speed on level ground, against a 32.0 N frictional force. He pushes in a direction 27.0° 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 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) Work done by friction=F dot S=32×17×cos 180=-544J

Since friction and displa3 are in opposite direction

b) Work done by gravitational force=F×17×cos90=0J

Since gravitational force is perpendicular to displacement

c) Work done by shopper=same as frictional work because friction is equal and opposite to applied force and it is given that body is not accelerating

But the force applied by shopper will be positive because the direction of the force applied by him and direction of displacement are same.

So work done by shopper=+544J

d) Let the force the shopper exerts be F

FScos27=544

F×17×cos 27=544 (using energy consideration by man)

F=36 N approx

e) The net force on the cart is zero because it is moving with constant speed. So total work done is zero on the cart.

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