6 replies to this topic

[Crimson Lego]

So been working on Grade 12 forces questions in-class, and ran into a little hump on the last few questions.

11. A boy drags his 80.0 N sled at a constant speed up a 15.0 º hill. He does so by pulling with a 35.0 N force on a rope attached to the sled. If the rope is inclined at 35.0 º to the horizontal;
a) what is the coefficient of kinetic friction between the sled and the snow?
b) At the top of the hill, he jumps on the sled (the boy weighs 50.0 kg) and slides down the hill. What is his acceleration down the slope?

13. A child pushes a 10.0 kg block up a 15.0 º ramp. She does so by exerting a horizontal force of 235.0 N. The coefficient of friction between the block and ramp is 0.23. The force is applied to the block over a distance of 5.0 cm. If the block was initially at rest, how far up the ramp will the block slide?



Thanks in advance to anyone who can help.

Edited by Leo Crimson, 13 February 2013 - 05:11 PM.

[Kisseena]

If I was back in high school, I would help, but I forgot everything. Maybe turtle can help.
What does E mean? Energy?

The answer is Q.

[Sir Turtlelot]

I could help you solve these, but I need to know what E stands for.

I think for 11-a, you need to use the formula F_f=µ_kN (Frictional Force = CoKF*Normal Force). This becomes more challenging if your object is on an incline.

I would need to go through my old notes for 11-b and 12.

[Crimson Lego]

I think Firefox just replaced degrees with E; must be the formatting or something. I'll just fix that.



And yeah, it was the incline that tripped me up; I'll give it another shot and see if I can use trig for that.

[Showsni]

Draw some pretty pictures! I'll go and scribble in paint until it looks right.

For the first bit, there's no acceleration, so all the forces should balance. You have an 80N force straight down, a 35N force at 35 degrees to the horizontal up the hill, a Normal force perpendicular to the hill and the frictional force will be acting down the hill. So, work out what that force is first.

(Picture one, if I can get paint to work)


So, let's think about everything in terms parallel to the hill. Then going down the hill we have mystery friction force plus some of that 80, and going up the hill we have some of that 35. Or, written out...

F + 80 sin 15 = 35 cos 20

So F = 35 cos 20-80 sin 15 = 12.1837181193N according to google calculator.

And perpendicular to the slope, we have some of that 80N = the Normal plus some of that 35N, or

80 cos 15 = N + 35 sin 20
N = 80 cos 15 - 35 sin 20 = 65.3033610867N

So, we know that F = mu N (How do you type mu? Well, friction coefficient, anyway.) So a quick substitution and rearranging gives mu = 0.18657107255

You probably want to put that to some significant figures.

Does that help you? Can you work out the rest?

Attached Files

•  sledphysics.jpg   19.7K   2 downloads

Edited by Showsni, 13 February 2013 - 05:42 PM.

[Crimson Lego]

I think I understand it better now, yeah; thanks a lot!

[arunma]

Well crap, you go and ask a physics question, and someone else beats me to it.