Question

A shopper in a supermarket pushes a cart with a force of 35.0 N directed at an angle of 25.08 below the horizontal. The force is just sufficient to balance various friction forces, so the cart moves at constant speed. (a) Find the work done by the shopper on the cart as she moves down a 50.0-m-long aisle. (b) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the friction force doesn’t change, would the shopper’s applied force be larger, smaller, or the same? (c) What about the work done on the cart by the shopper?

Answer 1

The work done by the shopper on the cart is calculated using the equation Work = Force x Displacement x cos(θ). If the shopper pushes the cart horizontally, her applied force will be larger, and the work done on the cart will also be greater.

Step-by-step explanation:

(a) To find the work done by the shopper on the cart as she moves down a 50.0-m-long aisle, we can calculate the work done using the equation:

Work = Force x Displacement x cos(θ)

Where the force is 35.0 N, the displacement is 50.0 m, and the angle (θ) is 25.08° below the horizontal. Plugging in these values, we get:

Work = 35.0 N x 50.0 m x cos(25.08°)

Calculating this will give us the work done by the shopper on the cart.

(b) If the shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before, the frictional force will remain the same. Therefore, the shopper's applied force will be larger to overcome both the frictional force and the vertical component of the gravitational force.

(c) The work done on the cart by the shopper will still be positive because she is still exerting a force on the cart in the direction of its displacement. However, the work done will be greater because the shopper now needs to exert more force.

Answer 2

a)1585 J

b)The applied force would be smaller and the work would again be zero

c)0 N

Step-by-step explanation:

a) The horizontal component of the applied force is 35 N x cos 25.08 = 31.70 N

Work = force x distance = 31.7 N x 50 m = 1,585 J

b) If the applied force were horizontal, all of it would counter friction, so only 31.7 N would be needed. The work done on the cart is zero because there is no change in mechanical energy.

c) The work done on the cart by the shopper will be zero because there is no change in mechanical energy.