Final answer:
The force exerted by the shopper is 33.0 N, as it offsets the work done by the 33.0 N frictional force over a distance of 25.0 m, resulting in constant speed of the grocery cart.
Step-by-step explanation:
The subject of the question relates to Physics, specifically involving force and work done by forces. If a shopper pushes a grocery cart 25.0 m at constant speed on level ground against a 33.0 N frictional force, we are to find the force the shopper exerts. Using energy considerations and the concept of work, we know that the work done by the frictional force is the product of the frictional force and the distance moved in the direction of the force. Since the speed is constant, the work done by the shopper must equal the work done against friction to maintain a non-accelerating (constant speed) condition.
The work done by friction is given by Work = force × distance, which is 33.0 N × 25.0 m = 825.0 J.
Because the cart is moving at constant speed, the net work done on the cart is zero.
Therefore, the force exerted by the shopper also results in 825.0 J of work, and hence, the force exerted by the shopper must be 33.0 N.