# Temal Energy

1. Mar 25, 2006

### PrimeTime89

Could some one possible help me with the problem i have for physics, it was a test question and i didn't get it at all.
A contestant will be required to pull a 50 Kg landing mat, from rest across the gym floor for 10 seconds. Mat has been selected for this event. he has calculated that if he could supply a constant force of 200N at an angle of 60 degrees to the horizontal, the speed of the mat after 10.0 seconds will be 8.00 m/s. hw also figured out during this time a constant frictional force acts on the mat.
a) what will be the change in Kinetic Energu of the mat during this 10 second period
b) how much work will mat do on the mat?
c) how much kenetic energy will be converted into energy heat during the 10 second interval?

It would be much apriciated if someone would be able to help me.
Thanks

2. Mar 25, 2006

### Kurdt

Staff Emeritus
If you could tell us what you have attempted for the problem thus far we would be able to help. A hint for the first part of the question to get you going is simply subtract the initial kinetic energy from the final kinetic energy.

3. Mar 25, 2006

### PrimeTime89

i just need to know how to get the avg potential and kenetic energy?
any help?

4. Mar 25, 2006

### Libertine

$$K.E. = \frac{1}{2}mv^2$$

5. Mar 25, 2006

### PrimeTime89

that's how you find the average potential gravitational and kenetic energies, then i am able to find the themal heat given off right, just to tell you guys i have done both a and b,i'm just stuck on c, wasn't quite sure how to do it

6. Mar 25, 2006

### Kurdt

Staff Emeritus
Work out the acceleration of the mat and what speed it should be after ten seconds and compare it with the actual speed. I hope this hint helps.

7. Mar 25, 2006

### PrimeTime89

didn't get it , so use, a=(V2+V1(t))/2, then v=d/t, then divide the 2 then use those finidings to calculate the EK and EG?

8. Mar 25, 2006

### Kurdt

Staff Emeritus
Work out the horizontal component of force. F=ma frm this equation you can work out the acceleration and then see how fast it should be going after 10 seconds. Compare the kinetic energy of this with the actual kinetic energy. The difference will be the energy wasted as heat.

9. Mar 25, 2006

### PrimeTime89

okay thank you very much

10. Mar 26, 2006

### PrimeTime89

so then he lost 600 N to thermal energy right?

11. Mar 26, 2006

### Kurdt

Staff Emeritus
No its considerably bigger and energy is measured in joules not newtons. If you show me your working I can tell you where you went wrong.

12. Mar 26, 2006

### PrimeTime89

okay
well i found the acceleration to be 4m/s2
then i found the distance to be 200m
then i found the speed to be 40m
but i think i did it wrong because i used the avg speed,
sdo i put in E.K. = .5(50)(20)2
and got 1000
then E.K. = .5(50)(4)2
then go 400
i think i know why it's wrong but not sure

13. Mar 26, 2006

### PrimeTime89

would it be 38400??

14. Mar 26, 2006

### Kurdt

Staff Emeritus
First of all the acceleration is given by Fcos(60) = ma because you only need the horizontal force pulling the mat. Secondly theres no need to find the distance just multiply the accelerat by the time (10 seconds) to find the speed it should be going if all the force went puerely into acceleration. Thirdly calculate the kinetic energy of the speed you find and the speed that it was actually going (8m/s) and take the difference of the two. No need to use average speed at all.

15. Mar 26, 2006

### PrimeTime89

is .25 the accel and is the 38400 J right or wrong?

16. Mar 26, 2006

### Kurdt

Staff Emeritus
38400 is wrong and the acceleration should be 2 m/s^2