> #work , #energy , #physics > A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0 below the horizontal. (a) What is the work done on the cart ?
Feb 2

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0 below the horizontal. (a) What is the work done on the cart ?

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0 below the horizontal. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the shopper? (c) Find the force the shopper exerts, using energy considerations. (d) What is the total work done on the cart?
1 view

Sort By
ChatGPT
Feb 2
The work done on an object is given by the equation:

Work = Force * Distance * cos(theta)

Where:
- Force is the applied force on the object
- Distance is the distance over which the force is applied
- theta is the angle between the force and the direction of motion

In this case, the applied force is the force with which the shopper pushes the cart, which we'll call F_applied. The distance over which the force is applied is 20.0 m. The angle between the force and the direction of motion is 25.0 degrees below the horizontal, which means it is 90 degrees - 25 degrees = 65 degrees with respect to the horizontal.

The frictional force is given as 35.0 N, but since it acts in the opposite direction of the applied force, we'll consider it as a negative force (-35.0 N).

Plugging in the values into the equation, we have:

Work = (F_applied - F_friction) * Distance * cos(theta)

Work = (F_applied - (-35.0 N)) * 20.0 m * cos(65 degrees)

Since the applied force is not given, we cannot calculate the exact work done on the cart.
0
×