Jump to content
Sal's RuneScape Forum

Physics questions


Recommended Posts

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 35.0º 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
Link to comment
Share on other sites

I don't quite have the time to respond now, but I will respond in full later unless others answered it. To help you on your way though, I recommend first drawing a diagram and knowing exactly what the forces are along the slope, ie. parallel to it - that's key for both questions.

For 11a, you'll want to figure out the force of friction and the force the boy exerts on the sled parallel to the slope, and then you can use those to figure out the coef. of friction. (knowing that the force of friction must be equal to the force the boy exerts on the sled because it's constant speed and thus F = m*a = m*0 = 0)

For 11b, I'm not quite as sure, but I think you'll want to remember your coefficient of friction, calculate the force exerted by gravity on the sled and the boy parallel to the slope, and then figure out the total force. Since you can figure out the mass from the given values, a = F/m.


For 13, I think you'll want to figure out the velocity of the block at the end of the 5 cm, then using the deceleration due to gravity figure out when velocity becomes 0, again remembering that you need the forces parallel to the slope.


Hope this helps!


EDIT: Here's the first question. I'll leave the second question to you, but if you have questions about them, ask.

11a. I'm skipping over the finding of the different components of the forces, I'm assuming they're no problem.

Ff = force of friction

Fp = force exerted by boy on rope

Fg = force of gravity,

N = normal force

u = coefficient of kinetic friction

'pa' appended to any variables simply refers to the component of the force parallel to the slope, 'pe' perpendicular.


Start out with what you know:

Ff + Fg,pa = Fp,pa

and Ff = uN

So: uN + Fg,pa = Fp,pa


Fg,pa is simply 80sin(15), Fp,pa = 35cos(20), and N = Fg,pe = 80cos(15)

Fill all that in, and you get

u*80cos(15) + 80sin(15) = 35cos(20)

u = (35cos(20)-80sin(15))/(80cos(15)) = 0,158


11b. For the sake of easy calculations, I'm taking g = 10 ms^-2


The total force exerted on the boy is Fg,pa - Ff (ie. gravity parallel to the slope accelerates him, but friction slows him down). This force, since he's accelerating, can also be written as F=ma, so


ma = Fg,pa - Ff


The mass of the boy is Mb = 50 kg, that of the sled is Ms = 80/10 = 8 kg, so m = Mb + Ms = 58 kg.

The total Fg is Fg,boy = 50*10 N and Fg,sled = 80 N, so Fg = 500+80 = 580N.


Fg,pa = 580sin(15)

Ff = uN = uFg,pe = 0,158*580cos(15)


So: a = (Fg,pa - Ff)/m = (580sin(15) - 0,158*580cos(15))/58 = 1,062 ms^-2


I think this is correct. The values seem fairly reasonable anyway.

Edited by reepicheep
Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

  • Create New...

Important Information

By using this site, you agree to our Guidelines and Privacy Policy.