Question

You place a m=2kg block at the bottom of a playground slide and give it quick kick, giving it an initial speed vi=12m/s. The slide is inclined at angle 35 degrees and the coefficient of kinetic friction between the block and the slide is mu_k=0.2. Use g=9.8m/s/s.

a) What is the magnitude of the friction for on the block while it is moving on the slide?
I found this to (correctly) be 3.2 N

b) Assume the slide is sufficiently long that the block does not go over the top. How far up the slide does the block go?
c) If the block slides back down, what will its speed be at the bottom?

Answer 1

Part b deals with the idea of conservation of energy.

Let me know if this helps.

Answer 2

can you help with my question?

Answer 3

I understand conservation of energy, but this problem is coming from an online course that has technically not introduced that topic yet. So I believe I have to find a way to solve it using what's been taught thus far, which is Newton's Laws/forces/kinematics/the current unit called Models of 1-D Motion

Answer 4

Just to see if it works I did try it using conservation of energy and got the wrong answer.
mgh = 1/2mv^2
(2)(9.8)h=1/2(2)(12)^2
7.3=h

For the ramp distance, sin35=7.3/x --> x=12.8
And that answer was wrong.

Answer 5

I'm trying to think about how to work the problem without the work energy theorem, but the final potential energy will be equal to the initial kinetic energy AND the work done by friction

Answer 6

That still didn't work, assuming I calculated it correctly.
mgh = 1/2mv^2 + Ff (should it be -Ff?)
(2)(9.8)h = 1/2(2)(12)^2 + 3.2
19.6h = 147.2
h= 7.5

sin35 = 7.5/x --> x = 13.1m

Answer 7

Ok, so I'm sorry for sending in the wrong direction with the conservation of energy idea.
Here's what you have going on

**It is important that you take into account that fact that because of the angle, the force of gravity now has both a y component and a x component.
**Also understand that frictional force will be opposite of the direction of movement.

You want to figure of the acceleration of the block which will then let you use your kinematic equations to solve for the distance the block moves up the slide. To do this you need to sum the forces on the block in the y direction.
\[\sum_{y}^{}F= N-mg \cos \theta\]
Because the block does not move up off of the slide at any point you can apply Newton's first law
\[N -mg \cos \theta =0 \rightarrow N =\cos \theta \]

Now summing the forces in the x direction \[\sum_{x}^{}F =mg \sin \theta +\mu N\]

The box is moving in the x direction and you can use Newton's second law which will allow you to solve for acceleration
\[mg \sin \theta + \mu N = ma\]

Again I apologize for misleading you at first. Hopefully this is more useful to you.

Answer 8

That worked and I got the correct distance. Thank you!

Answer 9

But now I'm having issues with part (c)...I keep getting 12 m/s which is obviously not correct.
Here are the givens I'm using: Vi=0, d=9.96, a=7.23, Vf=?

Answer 10

Figured it out. Didn't realize that although acceleration up and acceleration down are both constant, they have different values. Thanks so much for your help.