Answer:
See below.
Step-by-step explanation:
1. Newton's first law
(a) Problem 1
What is the net force required to keep a 500 kg object moving with a constant velocity of 10 m·s?
Answer: None.
Explanation: An object in motion stays in motion unless a net force acts on it.
The object will keep moving at 10 m/s.
(b) Problem 2
A force of 20 N acts on a 10 kg object from the left. A force of 30 N acts on it from the right. What is the net force required to keep the object moving with a constant velocity of 10 m/s?
Answer: 20 N
Explanation:
The net force is
30 N - 20 N = 10 N
You must apply another 10 N from the left. The net force is then:
30 N - 30 N = 0
If there is no net force, the object will keep moving at 10 m/s.
2. Newton's second law
(a) Problem 1
What is the net force needed to accelerate an object at a constant 5 m·s⁻²?
Answer: 5 N
Explanation:
F = ma = 1 kg × 5 m·s⁻² = 5 kg·m·s⁻² = 5 N
(a) Problem 2
A net force of 2 N acts on a 1 kg object. What is the magnitude and direction of the acceleration?
Answer: 2 m·s⁻²
Explanation:
F = ma
2 N = 1 kg × a
a = (2 N)/(1 kg) = (2 kg·m·s⁻²\1 kg) = 2 m·s⁻²
The direction of the acceleration is the same as that of the applied force.
2. Newton's third law
(a) Problem 1
A person with a mass of 58 kg is standing near you. Diagram and calculate the opposing forces.
Answer: 570 N up and down
Explanation:
See Fig. 1. The person's mass exerts a downward force on the floor.
F = mg, where g is the acceleration due to Earth's gravity
F = 58 kg × 9.8 m·s⁻² = 570 kg·m·s⁻² = 570 N
The floor exerts an upward force of 570 N.
(a) Problem 2
A teacher (mass 65 kg) pushes a cart (mass = 12 kg) of equipment (mass = 7 kg). Her foot applies a force of 150 N backward on the floor against a frictional force of 24 N. Diagram the opposing forces and calculate the net force available to move the cart.
Answer: 126 N
Explanation:
See Fig. 2 below. The teacher's mass exerts a downward force W on the floor, which exerts an equal reaction force R upward. The cart and equipment also exert a downward force on the floor, which exerts the same force upward. We can ignore these forces because they do not contribute to forward motion.
The teacher's foot exerts a backward force of 150 N on the floor, which exerts an equal force forward. However, a frictional force of 24 N opposes the forward force.
The net external force is the force of the floor minus the opposing frictional force. Thus,
F = 150 N − 24 N = 126 N