Newton's Laws Questions + Friction. Very tricky and help appreciated

[jambition]

I AM NOT INTERESTED IN DIRECT ANSWERS! I'm just stuck on how to process these questions and just need a little guidance on what I'm doing wrong. I will be forever indebted and I am more interested in learning how to solve for these. Thank you!


1. A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 2.6 m/s2.The acceleration of gravity is 9.8 m/s2. What is the magnitude of the force of the elevator floor on the person? Answer in units of k

1a. Relevant equations

f=m x a

1b. The attempt at a solution
Find mass of person (should be 900 N) but should I find the mass in kg?
Combine both accelerations and multiply by mass in kg?

2.When you drop a 0.35 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s2 toward the earth’s surface. According to Newton’s third law, the apple must exert an equal but opposite force on Earth.
If the mass of the Earth 5.98 × 1024 kg, what is the magnitude of the earth’s acceleration toward the apple?
Answer in units of m/s2.

2a. Relevant equations.
f= m x a?

2b. Attempt
The force the apple creates is 3.43. Do I set that equal to Eaarth's force and divide by Earth's mass? I get something like 5.74e-25. That can't be right.

3.The force of friction acting on a sliding crate is 232 N. How much force must be applied to maintain a constant velocity? Answer in units of N

3a. Equation
Not sure which one to use.

3b.Attempt
Do I find the mass of the crate? Acc. has to be 9.8 but how would i find for constant velocity?

 

[PhanthomJay]

I AM NOT INTERESTED IN DIRECT ANSWERS! I'm just stuck on how to process these questions and just need a little guidance on what I'm doing wrong. I will be forever indebted and I am more interested in learning how to solve for these. Thank you!


1. A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 2.6 m/s2.The acceleration of gravity is 9.8 m/s2. What is the magnitude of the force of the elevator floor on the person? Answer in units of k

1a. Relevant equations

f=m x a

It's f_net = m x a, where f_net is the algebraic sum of the force of gravity (the weight) on the person and the force of the floor (the normal force) on the person

1b. The attempt at a solution
Find mass of person (should be 900 N) but should I find the mass in kg?

the weight of the person is 900 N. The person's mass in kg is calculated by m=weight/g

Combine both accelerations and multiply by mass in kg?

No, find the net force acting on the person , and set it equal to m x a. The net force is always in the direction of the acceleration.

2.When you drop a 0.35 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s2 toward the earth’s surface. According to Newton’s third law, the apple must exert an equal but opposite force on Earth.
If the mass of the Earth 5.98 × 1024 kg, what is the magnitude of the earth’s acceleration toward the apple?
Answer in units of m/s2.

2a. Relevant equations.
f= m x a?

2b. Attempt
The force the apple creates is 3.43. Do I set that equal to Eaarth's force and divide by Earth's mass? I get something like 5.74e-25. That can't be right.

Why not? you don't expect the apple to lift up the Earth significantly, do you?

3.The force of friction acting on a sliding crate is 232 N. How much force must be applied to maintain a constant velocity? Answer in units of N

3a. Equation
Not sure which one to use.

3b.Attempt
Do I find the mass of the crate? Acc. has to be 9.8 but how would i find for constant velocity?

This problem assumes the crate is on a rough horizontal surface. If it's moving at constant velocity, there is NO acceleration (per Newton 1). So what force must be applied if it is moving at constant velocity?

Welcome to Physics Forums!

 

[jambition]

Ah thank you! I think I'm getting somewhere now! Problem 2 was correct after all

For problem 3, I'm still not understanding what I'm looking for exactly. The follow up question to that problem asks for a net force so I must be looking for some kind of acceleration, no?

 

[PhanthomJay]

For problem 3, I'm still not understanding what I'm looking for exactly. The follow up question to that problem asks for a net force so I must be looking for some kind of acceleration, no?

Not if the net force is zero, that's Newton 1 'An object in motion at constant speed moving in a straight line will remain at constant speed in a straight line unless acted on by a net unbalanced force. No net force, then no acceleration, F_net =0.

 

[rwisz]

1a. Relevant equations: F = ma (Newton's Second Law)

1b. The attempt at a solution: To find the magnitude of the force of the elevator floor on the person, we can use Newton's Second Law. First, we need to convert the person's weight from kN to N. 0.9 kN is equal to 900 N. Then, we can use the combined acceleration of gravity and the elevator, which is 9.8 m/s^2 + 2.6 m/s^2 = 12.4 m/s^2. Finally, we multiply the mass (in kg) by the combined acceleration to get the force in Newtons. So, the equation would be F = (900 N) x (12.4 m/s^2) = 11160 N = 11.16 kN. Therefore, the magnitude of the force of the elevator floor on the person is 11.16 kN.

2a. Relevant equations: F = ma (Newton's Second Law)

2b. Attempt: To find the magnitude of the Earth's acceleration toward the apple, we can use Newton's Second Law again. First, we need to convert the mass of the Earth from kg to N. 5.98 x 10^24 kg is equal to 5.98 x 10^27 N. Then, we can use the force exerted by the apple (3.43 N) and divide it by the mass of the Earth in N to get the acceleration in m/s^2. So, the equation would be a = (3.43 N) / (5.98 x 10^27 N) = 5.74 x 10^-25 m/s^2. This is a very small acceleration, but it is correct.

3a. Relevant equations: F = ma (Newton's Second Law)

3b. Attempt: To find the force needed to maintain a constant velocity, we need to use Newton's Second Law again. However, in this case, the acceleration is 0 m/s^2 since the velocity is constant. So, the equation would be F = (232 N) x (0 m/s^2) = 0 N. Therefore, no additional force is needed to maintain a constant velocity. The force of friction is equal and opposite to the force applied to the crate, so if the crate is already moving at a