A shopper pushes a shopping cart with a force of 35N directed at an angle of 25

Question

A shopper pushes a shopping cart with a force of 35N directed at an angle of 25 degrees below the horizontal. The cart moves at a constant speed.

A) Find the work done by the shopper as she moves down a 50.0 m length aisle.

B) what is the net work done on the cart? Why?

C) The shopper goes down the next aisle pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn't change, whould the shopper's applied force be larger or smaller or the same? What about the work done on the cart by the shopper?

Explanation / Answer

(A) horizontal componant of force = F cos theta = 35 x .9 = 31.5

W = F.s = 31.5 x 50 = 1575 J

(B) here V = constant means a = 0 means net force on cart = 0 means horizontal componant = friction

Net W = Fnet x S = 0 x 50 = 0 J

(Because the net F = 0)

(C)Here the cart going down means an gravitational force in downward direction is comes here so as velocity don't change means

F + Fg = friction

So here an extra Force Fg in the direction of F so the magnitude of F should decreases to maintain the forces equal.

As F is smaller now so W by shopper also decreases.

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