Horizontal Spring Problem, just need it checked!

Becca_Gadd

A 3kg block is resting at the bottom of a ramp that is at a 37 degree angle with a coefficient of friction of .2. A spring at the bottom of the hill has a spring constant of 5000N/m. The block compresses the spring 8cm (so .08m) and is then released. The block slides up the ramp and then slides back down the ramp.

a. Find the velocity of the block as it starts to slide up the ramp.
I did this problem, and my teacher said it was right. V = 3.26m/s

b. Find the velocity of the block when it returns to the bottom of the ramp on the way back down.
I got 3.38 m/s, is this right?

Ff =16*2=3.2N 16-3.2=12.8 12.8=28l l=.711
WFf=Fd W=(3.2)(.711)=2.21(2)=4.55
16-4.55=11.45J KE=1/2m^2 11.45=.5(2)v^2 v=3.38 m/2

c. Find the amount that the block will compress the spring when it comes back down.
Is it asking how many joules or how many meters? I got .067m, is this right?

11.45=.5(5000)x^2 11.45=2500x^2
.00458=x^2 x=.067m

Note: This isn't homework. It's not a grade of any sort. I'm doing some practice problems from old worksheets to prepare for a test tomorrow. So I'm trying to understand how to do it, the concepts of it, not just what these answers are. I could care less.

Becca_Gadd

I'm going to fail this test.

ehild

Show a picture and explain your notations, please.

ehild

Becca_Gadd

This is at the beginning of the problem.

a. I know it's right, no need to explain.

b. Find the velocity of the block when it returns to the bottom of the ramp on the way back down.
I got 3.38 m/s, is this right?

I used the equal Ff = FnU (force of friction = Force up * coefficient of friction) =16*.2=3.2N

I think subtraced the force of friction from the total kinetic energy 16-3.2=12.8 I think took this answer and plugged it into this equation to find the length of the ramp. 12.8=28l l=.711
The work of the force of friction = force *distance. WFf=Fd W=(3.2)(.711)=2.21 So I got the force of work for the way up. And then multiplied by two because there's friction on the way down as well. 2.21(2)=4.55 The total force of friction is 4.55J
I subtracted that from the total energy to get the ttoal work. 16-4.55=11.45J I then used this equation to get the kinetic energy of the box at the bottom of the ramp: KE=1/2m^2 And plugged in the numbers: 11.45=.5(2)v^2 And simplified: v=3.38 m/2

c. Find the amount that the block will compress the spring when it comes back down.
Is it asking how many joules or how many meters? I got .067m, is this right?
I used the equal (potential spring energy) Eps = 1/2kx^2 (1/2 spring constant times how far it was stretched squared) And plugged in my numbers 11.45=.5(5000)x^2 And solved for x
11.45=2500x^2 .00458=x^2 x=.067m

ehild

The block first moves horizontally, accelerated by the spring while it has some final velocity. After detached from the spring it starts moving up on the slope. The force of friction acts on the slope, it is the normal force between the slope and block multiplied by the coefficient of friction. What do you mean on "force up"? Where is "16" come from?

ehild

ehild

I have found out that 16 J is the initial KE of the block. It is not force!! You can not add or subtract force and energy.

Use the angle of the slope to find the normal force.

ehild

Becca_Gadd

By "force up" I mean "normal force." That's just what my teacher refers to it as. The force is applied UPWARDS. It's the slope pushing back on the box, which is UP, so meh. And I already calculated that later on, 30sin37=18 N.
I don't see how that applies to the first part. I plugged it in where I had 16N, just to see if that might be what you mean, but it's not. I obviously don't know how to start the beginning of this, but I know how to do the rest. What equation are you supposed to use at the beginning? I'm sorry I'm doing this so unconventionally, my teacher teaches from his head, and thus the only examples we have are the ones we make & figure out ourselves. So telling me to "think about it" isn't going to do anything, because I never learned. I need to subtract the friction from something, but I'm not sure what. I thought I was supposed to subtract it from the total work, but I guess I can't do that. Should I convert the friction to work and then subtract that? I don't know. Edit: nope, tried that, that wasn't it.

ehild

You answered question a, it is "b" now. The block gained kinetic energy KE=16 J from the spring, and goes up the slope. During this, it loses KE because of gravity and friction. But the force of friction is u*30*cos(37). You need to find out how far upward the block slides on the slope till it stops.

ehild

ehild

Take care. The component of gravity along the slope is mgsin(θ), it points downward. The normal force "out of the slope" is mgcos(θ), so the force of friction is umgcos(θ). When moving upward, both forces act downward, against the motion of the block. It will stop when the work of both forces consumes all the initial KE of the block. L(mgsin(θ) +umgcos(θ)) = KE(=16 J). Find L.

ehild

Becca_Gadd

Ah, thank you! L(30)(.6)+(.2)(30)(.8)=16 L=.702, that's the answer I had written down from class. I'll try and figure out the rest with this new solution, thanks!