Grade 11 physics
Assignment 5 Newton’s Laws
Please complete each question in “given/formula(s)/solution/answer” format.
An elevator has a mass of 2250 kg.
a) What is the tension in the cable that supports the elevator when it is at rest? Complete the FBD
b) What is the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s2? Complete the FBD.
c) What is the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s2? Complete the FBD.
a) Given:
Mass of elevator (m) = 2250 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Formula:
Tension (T) = m * g
Solution:
T = 2250 kg * 9.8 m/s^2
T = 22050 N
Answer:
The tension in the cable that supports the elevator when it is at rest is 22050 N.
b) Given:
Mass of elevator (m) = 2250 kg
Acceleration (a) = 0.55 m/s^2 (upwards)
Formula:
Tension (T) = m * (g + a)
Solution:
T = 2250 kg * (9.8 m/s^2 + 0.55 m/s^2)
T = 2250 kg * 10.35 m/s^2
T ≈ 23287.5 N
Answer:
The tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2 is approximately 23287.5 N.
c) Given:
Mass of elevator (m) = 2250 kg
Acceleration (a) = 0.55 m/s^2 (downwards)
Formula:
Tension (T) = m * (g - a)
Solution:
T = 2250 kg * (9.8 m/s^2 - 0.55 m/s^2)
T = 2250 kg * 9.25 m/s^2
T ≈ 20812.5 N
Answer:
The tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 is approximately 20812.5 N.
a) FBD:
- The force of gravity acting on the elevator is given by F_g = m * g, where m is the mass of the elevator (2250 kg) and g is the acceleration due to gravity (9.8 m/s^2).
- The tension in the cable that supports the elevator when it is at rest is equal to the force of gravity, so Tension = F_g = m * g.
Given:
m = 2250 kg
g = 9.8 m/s^2
Solution:
Tension = m * g
Tension = 2250 kg * 9.8 m/s^2
Tension = 22050 N
Answer:
The tension in the cable that supports the elevator when it is at rest is 22050 N.
b) FBD:
- The force of gravity acting on the elevator is still F_g = m * g.
- The net force acting on the elevator when it is moving up with an acceleration of 0.55 m/s^2 is given by F_net = m * a, where a is the acceleration (0.55 m/s^2).
- The tension in the cable is equal to the force of gravity plus the net force, so Tension = F_g + F_net.
Given:
m = 2250 kg
g = 9.8 m/s^2
a = 0.55 m/s^2
Solution:
Tension = F_g + F_net
Tension = m * g + m * a
Tension = 2250 kg * 9.8 m/s^2 + 2250 kg * 0.55 m/s^2
Tension = 22050 N + 1237.5 N
Tension = 23287.5 N
Answer:
The tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2 is 23287.5 N.
c) FBD:
- The force of gravity acting on the elevator is still F_g = m * g.
- The net force acting on the elevator when it is moving down with an acceleration of 0.55 m/s^2 is given by F_net = m * a. Since it is moving down, the acceleration is in the opposite direction, so the net force is negative.
- The tension in the cable is equal to the force of gravity minus the net force, so Tension = F_g - F_net.
Given:
m = 2250 kg
g = 9.8 m/s^2
a = 0.55 m/s^2
Solution:
Tension = F_g - F_net
Tension = m * g - m * a
Tension = 2250 kg * 9.8 m/s^2 - 2250 kg * 0.55 m/s^2
Tension = 22050 N - 1237.5 N
Tension = 20812.5 N
Answer:
The tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 is 20812.5 N.
To solve these questions, we need to apply Newton's laws of motion. Let's start with some background information:
1. Newton's First Law: An object at rest tends to stay at rest, and an object in motion tends to stay in motion with the same speed and direction, unless acted upon by an external force.
2. Newton's Second Law: The net force acting on an object is equal to the product of its mass and acceleration. Mathematically, this can be represented by the formula F = ma, where F is the force, m is the mass, and a is the acceleration.
3. Newton's Third Law: For every action, there is an equal and opposite reaction.
Now let's proceed to answer the questions:
a) What is the tension in the cable that supports the elevator when it is at rest?
When the elevator is at rest, the net force acting on it is zero because it is not accelerating. Therefore, the tension in the cable should be equal to the weight of the elevator.
Given that the mass of the elevator is 2250 kg (m = 2250 kg), we can calculate the weight using the formula weight = mass × acceleration due to gravity (g).
The acceleration due to gravity on Earth is approximately 9.8 m/s^2.
So, weight = m × g = (2250 kg) × (9.8 m/s^2) = 22050 N.
Hence, the tension in the cable when the elevator is at rest is 22050 N.
b) What is the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2?
When the elevator is moving up with an acceleration of 0.55 m/s^2, we need to consider the additional force due to this acceleration.
In this case, the net force acting on the elevator is the sum of the tension force (T) and the force due to the elevator's mass and acceleration (ma).
Using Newton's second law (F = ma), we can write the equation:
T + ma = m × (0.55 m/s^2),
where T is the tension in the cable.
Rearranging the equation, we get:
T = m × (0.55 m/s^2) - ma.
Plugging in the given mass (m = 2250 kg) and the acceleration (0.55 m/s^2), we can calculate T.
T = (2250 kg) × (0.55 m/s^2) - (2250 kg) × (9.8 m/s^2).
Simplifying the calculation, we find that the tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2 is approximately 3945 N.
c) What is the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2?
When the elevator is moving down with an acceleration of 0.55 m/s^2, we again need to consider the additional force due to this acceleration.
Similar to the previous question, the net force acting on the elevator is the sum of the tension force (T) and the force due to the elevator's mass and acceleration (ma).
Using Newton's second law (F = ma), we can write the equation:
T - ma = m × (0.55 m/s^2).
Again, rearranging the equation, we get:
T = m × (0.55 m/s^2) + ma.
Plugging in the given mass (m = 2250 kg) and the acceleration (0.55 m/s^2), we can calculate T.
T = (2250 kg) × (0.55 m/s^2) + (2250 kg) × (9.8 m/s^2).
Simplifying the calculation, we find that the tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 is approximately 2992 N.
So, the answers are:
a) Tension in the cable when the elevator is at rest = 22050 N.
b) Tension in the cable when the elevator is moving up with an acceleration of 0.55 m/s^2 = 3945 N.
c) Tension in the cable when the elevator is moving down with an acceleration of 0.55 m/s^2 = 2992 N.