A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of 22.0 m. She pushes it in the direction 30.0° below the horizontal. A 48.0-N frictional force opposes the motion of the cart. (a) Draw the free-body diagram showing the forces that act on the cart. (b) What is the magnitude of the force that the shopper exerts? Determine the work done by (c) the pushing force, (d) the frictional force, and (e) the gravitational force.
The Correct Answer and Explanation is :
Solution:
Let’s break the problem down step by step.
(a) Free-Body Diagram
The forces acting on the shopping cart are:
- Gravitational Force ((F_g)): Acts downward ((mg)).
- Normal Force ((F_N)): Acts upward, exerted by the ground.
- Frictional Force ((F_f)): Opposes motion, acting leftward.
- Applied Force ((F_A)): Exerted by the shopper at an angle (30^\circ) below the horizontal.
(b) Magnitude of the Force Exerted by the Shopper
Since the cart moves at a constant velocity, the horizontal component of the applied force must balance the frictional force. That is:
[
F_A \cos 30^\circ = F_f
]
Substituting (F_f = 48.0 N):
[
F_A \cos 30^\circ = 48.0
]
[
F_A = \frac{48.0}{\cos 30^\circ}
]
[
F_A = \frac{48.0}{0.866}
]
[
F_A \approx 55.4 N
]
(c) Work Done by the Pushing Force
Work is given by:
[
W = F d \cos \theta
]
[
W_A = (55.4)(22.0) \cos 30^\circ
]
[
W_A = (55.4)(22.0)(0.866)
]
[
W_A \approx 1054 J
]
(d) Work Done by the Frictional Force
Since friction acts opposite to the displacement, the angle is (180^\circ), so:
[
W_f = F_f d \cos 180^\circ
]
[
W_f = (48.0)(22.0)(-1)
]
[
W_f = -1056 J
]
(e) Work Done by the Gravitational Force
Since gravity acts perpendicular to the displacement, the work done is:
[
W_g = mg d \cos 90^\circ = 0 J
]
Explanation (300 Words)
In this problem, we analyze the forces acting on a shopping cart being pushed at an angle while moving at constant velocity. The key concept is force decomposition. Since the cart is moving at constant velocity, the net force in the horizontal direction is zero, meaning the applied horizontal force must balance the opposing frictional force.
To determine the force exerted by the shopper, we resolve it into horizontal and vertical components. The horizontal component must counteract friction, allowing us to solve for the total applied force. Using trigonometry, we find the shopper exerts about 55.4 N.
Next, we calculate work done by different forces. Work is defined as the force applied in the direction of displacement times the distance moved. The shopper’s applied force contributes positively to the work done, with a magnitude of 1054 J. Friction opposes motion, so it does negative work, equal to -1056 J. The gravitational force acts perpendicular to the displacement, meaning it does zero work.
This analysis shows how forces interact in everyday situations like pushing a shopping cart. Since the forces balance in the horizontal direction, the cart does not accelerate. Understanding these principles is essential in physics, especially Newton’s laws of motion and the work-energy theorem.