# Homework Help: Is anyone here familiar with the program masterphysics?

1. Oct 12, 2011

1. The problem statement, all variables and given/known data

Ok, so masteringphysics has rejected an answer I think is right. The question is set up with "insert value" and then "insert units"

Heres the question.

A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t) = (2.80 m/s)t +(0.61 m/s^3)t^3

What is the magnitude of the force F when 3.60 s?

F= (value)(units)

I inserted 65.9 and N

It rejected it.

I tried 65.9 and kg*m/s^2

Rejected again.

I have one try left, and I don't understand what is wrong. Thanks.

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 12, 2011
2. Oct 12, 2011

### bbbeard

Maybe you should put in all the decimal places (65.88), not just the significant digits.

3. Oct 12, 2011

Usually if my answer is to the wrong amount of digits, it will tell me in the feedback.

4. Oct 12, 2011

### Liquidxlax

Lol i had that same question 3 years ago, sorry i can't remember what i entered, but i know the answer is 65.99N. I'd say talk to you prof/teacher before you use up all your chances. That is the major flaw in those internet based assignments like masteringphyics and masteringchemsitry.

5. Oct 12, 2011

Haha, 3 years and they still didn't fix it. Alright, I will send my prof. an e-mail.

Its a very frustrating flaw.

6. Oct 12, 2011

### Staff: Mentor

Realize that you are asked to find the force F, which is not the net force.

7. Oct 12, 2011

Do I just plug 3.66 into the original equation for the answer of 40.2 N?

8. Oct 12, 2011

### Staff: Mentor

No.

Draw a Free Body Diagram and set up Newton's 2nd law. What's the acceleration at the specified time?

9. Oct 12, 2011

F = ma

acceleration is 13.176 m/s^2 (3.66*3.60)

m = 5.00

F = 65.9 N?

10. Oct 12, 2011

### Staff: Mentor

That should be ƩF = ma.

What forces act on the crate? The applied force 'F' is just one of the forces acting.

Good.

No, that's the net force.

11. Oct 12, 2011

Would tension be a factor?

12. Oct 12, 2011

### Staff: Mentor

The applied force 'F' is the tension. Recall that it's "applied to the end of the rope".

What other force acts?

13. Oct 12, 2011

Weight(49 N)
Normal force
65.88 N - 49 N = 16.88 N?

Last edited: Oct 12, 2011
14. Oct 12, 2011

### Staff: Mentor

Right.
No.

Do it systematically:

What forces act? (One of them will be the force 'F' which you're solving for.)
Write an expression for the net force.
Set the net force equal to 'ma'.
Solve for the applied force F.

15. Oct 12, 2011

Ok I will try to figure this out. Thanks!

16. Oct 12, 2011

### Staff: Mentor

You're almost there. You just need to organize your approach. Don't give up!

17. Oct 12, 2011

T-W=ma
T-49 N = 65.88 N
T = Fapplied = 114.88 N

18. Oct 12, 2011

### Staff: Mentor

Perfecto!

19. Oct 12, 2011