A shopper in a supermarket pushes a cart with a force of 35N directed at an angl

Question

A shopper in a supermarket pushes a cart with a force of 35N directed at an angle of 25 degrees below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at a constant speed. (a) Find the work done by the shopper as she moves down a 50 m length aisle. (b) what is the net work done on the cart? (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 chagne, would the shopper's applied force be larger, smaller or the same? What about the work done on the cart by the shopper?

Explanation / Answer

A shopper in a supermarket pushes a cart with a force of 35 N directed at an angle of 25° below the horizontal?

Work = Force * distance
Since the distance is horizontal, you need the horizontal component of the force.
Horizontal component = 35 * cos 25

Work done by shopper = 35 * cos 25 * 50

The friction force is horizontal.
Since the force is just sufficient to balance various friction forces, so the cart moves at constant speed, the magnitude of the total friction force must equal the magnitude of the horizontal component of the force of the shopper. The direction of the total friction force is opposite the direction of motion.

So the friction force = -35 * cos 25

The net force on the cart = 35 * cos 25 + -35 * cos 25 = 0
The net work done on the cart by all forces = 0 * 50 = 0

OR

Net force = total mass * acceleration
Since the cart moves at constant speed, acceleration = 0 m/s^2
Net force = total mass * 0
So net force = 0
The net work done on the cart by all forces = 0 * 50 = 0

c)

Horizontal component of force = 35 cos(25) = 31.72 N
Vertical component of force = 35 sin(25) = 14.79 N

If the shopper now goes down the aisle pushing horizontally, he or she will only need to apply a force of 31.72 N to overcome frictional forces.

Work done = force * distance moved in the direction of the force.

Horizontally, that is 31.72 * distance moved in both cases.

In the first part of the question, the vertical component of the force is matched by the reaction force of the floor on the cart, so the cart does not move vertically.

Work done vertically = 14.79 * 0 = 0 J

So overall, the work done is the same in both cases

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