A person pushes a 16.0-kg shopping cart, setting the stage for an exploration of forces, motion, work, and energy transformations. This detailed analysis delves into the intricate interplay of physical principles that govern the cart’s behavior, providing a comprehensive understanding of its dynamics.
Person Pushing a 16.0-kg Shopping Cart: A Person Pushes A 16.0-kg Shopping Cart
A person pushes a 16.0-kg shopping cart with a force of 25.0 N at an angle of 30.0° below the horizontal. The cart starts from rest and moves a distance of 10.0 m.
Forces Acting on the Shopping Cart
The forces acting on the shopping cart are:
- Weight (mg):The force of gravity acting downward on the cart, equal to 16.0 kg – 9.8 m/s² = 156.8 N.
- Normal force (N):The upward force exerted by the floor on the cart, equal in magnitude to the weight.
- Applied force (F):The force applied by the person, equal to 25.0 N at 30.0° below the horizontal.
- Friction (f):The force opposing the motion of the cart, acting parallel to the floor.
A diagram of the forces acting on the cart is shown below:
Motion of the Shopping Cart, A person pushes a 16.0-kg shopping cart
The applied force causes the cart to accelerate. The acceleration is given by:
a = F / m
where F is the applied force, m is the mass of the cart, and a is the acceleration.
Plugging in the values, we get:
a = 25.0 N / 16.0 kg = 1.56 m/s²
The cart starts from rest, so its initial velocity is 0 m/s. The final velocity of the cart is given by:
v² = u² + 2as
where u is the initial velocity, v is the final velocity, a is the acceleration, and s is the distance traveled.
Plugging in the values, we get:
v² = 0² + 2 – 1.56 m/s² – 10.0 m
v = 5.66 m/s
Therefore, the final velocity of the cart is 5.66 m/s.
Work Done on the Shopping Cart
The work done on the shopping cart by the person pushing it is given by:
W = F – d – cos(θ)
where W is the work done, F is the applied force, d is the distance traveled, and θ is the angle between the force and the displacement.
Plugging in the values, we get:
W = 25.0 N – 10.0 m – cos(30.0°) = 216.5 J
Therefore, the work done on the cart is 216.5 J.
Energy Changes in the System
The energy changes that occur in the system as the person pushes the shopping cart are:
- Kinetic energy:The kinetic energy of the cart increases as it accelerates.
- Potential energy:The potential energy of the cart remains constant because the cart is moving on a level surface.
- Work:The work done by the person on the cart is converted into kinetic energy.
A table of the energy changes is shown below:
| Energy | Initial | Final |
|---|---|---|
| Kinetic energy | 0 J | 108.25 J |
| Potential energy | 1568 J | 1568 J |
| Work | 0 J | 216.5 J |
Helpful Answers
What is the primary force acting on the shopping cart?
The primary force acting on the shopping cart is the force applied by the person pushing it.
How does the force applied by the person affect the cart’s motion?
The force applied by the person accelerates the cart, causing it to move.
What is the work done on the shopping cart?
The work done on the shopping cart is the force applied by the person multiplied by the distance the cart moves.
