Question

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 25.0º below the horizontal.

(a) What is the work done on the cart by friction?
a) 0 , {J}
b) -700 , {J}
c) 700 , {J}
d) 1400 , {J}

Answer 1

The work done on the cart by friction is 700 J. The work done on the cart by the gravitational force is zero. The work done on the cart by the shopper is also zero.

Step-by-step explanation:

(a) The work done on the cart by friction can be calculated using the formula:

Work = Force x Distance x cos(theta)

where the force is the frictional force and the distance is the distance the cart is pushed. The angle theta is the angle between the force and the direction of motion, which in this case is 25.0° below the horizontal.

So, the work done on the cart by friction is:

Work = 35.0 N x 20.0 m x cos(25.0°) ≈ 700 J

(b) The work done on the cart by the gravitational force is zero because the cart is pushed horizontally and the gravitational force acts vertically.

(c) The work done on the cart by the shopper is equal to the work done on the cart by friction, which is 700 J.

(d) To find the force the shopper exerts, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Since the cart is pushed at a constant speed, its kinetic energy doesn't change, so the work done on the cart by the shopper is zero. Therefore, the force the shopper exerts is also zero.

(e) The total work done on the cart is the sum of the work done by friction and the work done by the shopper, which is 700 J + 0 J = 700 J.