A shopper pushes a grocery cart for a distance 21.5 m at a constant speed on level ground, against a 45 N frictional force. He pushes in a direction 27.5° below the horizontal.a.) What is the work done on the cart by friction, in joules? b.) Find the magnitude of the force, in newtons, that the shopper exerts on the cart.c.) What is the total work done on the cart, in joules?

Question
Asked Oct 28, 2019
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A shopper pushes a grocery cart for a distance 21.5 m at a constant speed on level ground, against a 45 N frictional force. He pushes in a direction 27.5° below the horizontal.

a.) What is the work done on the cart by friction, in joules?

b.) Find the magnitude of the force, in newtons, that the shopper exerts on the cart.

c.) What is the total work done on the cart, in joules?

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Expert Answer

Step 1

Given:

Displacement of the cart = 21.5 m                    

Friction force = 45 N

Direction of the force =  27.5° below the horizontal

Step 2

(a) Calculating the work done on the car by friction:

Force of friction = 45 N
Displacement of the block 21.5 m
(45N) x (21.5m)=-967.5J
Work done
Negative sign shows that the force and displacement are in opposite direction
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Force of friction = 45 N Displacement of the block 21.5 m (45N) x (21.5m)=-967.5J Work done Negative sign shows that the force and displacement are in opposite direction

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Step 3

(b) Calculating the magnitude of force th...

As the cart moves with a constant velocity, hence acceleration of the cart must be zero or the
net force on the car must be zero.
So, the shopper must exert such force which produces a horizontal component equal to the
fricton force on the block.
Let the force exerted by the shopper is F at an angle 27.5 degree below the horizontal
Hence, F cos(27.5°) 45N
=
F = 50.73N
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Image Transcriptionclose

As the cart moves with a constant velocity, hence acceleration of the cart must be zero or the net force on the car must be zero. So, the shopper must exert such force which produces a horizontal component equal to the fricton force on the block. Let the force exerted by the shopper is F at an angle 27.5 degree below the horizontal Hence, F cos(27.5°) 45N = F = 50.73N

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