# Block, spring, and incline- work/energy

1. Jul 10, 2013

### natasha13100

1. The problem statement, all variables and given/known data
A small block of mass 0.8 kg is launched by a compressed spring with force constant k=400 N/m. The initial compression of the spring is 0.12 m. The block slides along a horizontal frictionless surface and then up an inclined plane that makes an angle θ=30° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is μk=.2. (See the attached picture.) Find the maximum vertical height h reached by the block.

2. Relevant equations
Uspr=potential energy of a spring=kx2/2 where k is the spring constant
K=kinetic energy=1/2mv2
Ug=potential energy due to gravity=mgh
G=gravitational force=mg where g=force due to gravity=-9.8m/s2
Ff=force due to friction (kinetic)=μkN
Ftot=total force=ƩF
F=force=ma
v2=vi2+2ax where v=velocity, a=acceleration, and vi=initial velocity
other equations may be used

3. The attempt at a solution
I need help understanding where I went wrong and how the problem should be done.
Uspr=kx2/2=(400)(.12)2/2=2.88
K=Uspr
K=1/2mv2=1/2(.8)v2=2.88
v0=3√(2)/5
I assume right is in the +x direction and up is in the +y direction.
G=-7.84
G=-Ny=-Ncos(θ)
N=-G/cos(θ)=7.84/cos(30°)
Ff=-μkN=-.2*7.84/cos(30°)=1.568/cos(30°)
Ftoty=ƩFy=7.84/cos(30°)-7.84-1.568/cos(30°)=-.5977
v0y=3√(2)cos(30°)/5=3√(6)/10
Ftoty=may
ay=Ftoty/m=-.5977/.8=-.747125
vy2=v0y2+2ayh
vy2=0 when the block is at its maximum height
0=(3√(6)/10)2+2(-.747125)h
1.49425h=5.4
h≈3.613853104

#### Attached Files:

• ###### block, spring, and incline work and energy.png
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Last edited: Jul 10, 2013
2. Jul 10, 2013

### TSny

Welcome to PF!

Check what you wrote for the potential energy of the spring.

For objects accelerating along an incline, you will be much better off choosing one of your axes along the incline and the other axis perpendicular to the incline. That way, one of your axes will be parallel to the acceleration vector of the object.

3. Jul 16, 2013

### natasha13100

Ok so I'm still doing something wrong. Here are some of my latest thoughts on the problem:
Uspr=kx2/2
Due to conservation of energy, Uspr=K
K=mv2/2
kx2/2=mv2/2 so v=√(kx2/m)
I use the coordinate where the direction of the normal force is +y and up the incline is +x.
Fy=0 since the block is on the incline.
Fy=N-Gy since N and Gy are in opposite directions.
Gy=mg/cosθ so N=mg/cosθ
Fx=Ff+Gx since Ff and Gxare in the same direction
FfkN=μkmg/cosθ and Gx=mg/sinθ
therefore, Fxkmg/cosθ+mg/sinθ.
I've tried multiple approaches here:

1.F=Fx=ma so a=Fx/m=μkg/cosθ+g/sinθ
vf2=0 because the block isn't moving when it reaches its highest point of the incline and vi2=kx2/m from earlier. vf2=vi2+2ad so 0=kx2/m+2(μkg/cosθ+g/sinθ)d and d=kx2/(2mg*(μk/cosθ+1/sinθ)). sinθ=h/d so h=dsinθ=kx2sinθ/(2mg*(μk/cosθ+1/sinθ)).

2. w=Fcosθx so w=-Fxd=-mgd(μk/cosθ+1/sinθ)
Ei=Ef+W
K=Ug+W
K=Uspr=kx2/2, Ug=mgh
kx2/2=mgh-mgd(μk/cosθ+1/sinθ)
*since d=h/sinθ, kx2/2=mgh-mgh(μk/cosθ+1/sinθ)/sinθ
therefore, h=kx2sinθ/(2mg(μktanθ+cotθ))

3. This try combines the two above.
Instead of doing d=h/sinθ(* in 2), I used the d I got in 1 and came up with mgh=kx2/2+kx2/2+kx2 so h=kx2/(mg)

I have no idea where I'm going wrong or what else to try.

4. Jul 16, 2013

### TSny

The main problem seems to be in your expressions for the x and y components of the gravitational force Gx and Gy on the incline. Check your notes for other inclined plane problems.

Your method 1 looks good except for the expressions for Gx and Gy.

Your method 2 is incorrect because you have double-counted the effect of gravity. The work that gravity does as the block goes up the incline is equal in magnitude to the change in Ugrav. If you include both the work done by gravity and the change in U, then you are essentially including the work done by gravity twice.

[EDIT: Also, there's a mistake in how you arranged the terms when you wrote Ei=Ef+W. Can you spot it?]

5. Jul 16, 2013

### TSny

On the slope, is Fx positive or negative based on your direction for the positive x-axis?

So, should the acceleration be positive or negative as the block goes up the slope?

6. Jul 16, 2013

### natasha13100

Is this correct?
Gx=-mgsinθ and Gy=-mgcosθ
The force and therefore the acceleration are negative.

7. Jul 17, 2013

### TSny

Yes. That looks good.

8. Jul 17, 2013

### natasha13100

Thank you so much!
So that means N=-mgcosθ
Fx=Ff+Gx
Ff=μkN=-μkmgcosθ and Gx=-mgsinθ
therefore, Fx=-μkmgcosθ-mgsinθ.
F=Fx=ma so a=Fx/m=-μkgcosθ-gsinθ
vf2=0
vf2=vi2+2ad so 0=kx2/m-2(μkgcosθ+gsinθ)d and d=kx2/(2mg*(μkcosθ+1sinθ)). sinθ=h/d so h=dsinθ=kx2sinθ/(2mg*(μkcosθ+1sinθ)).

Last edited: Jul 17, 2013
9. Jul 17, 2013

### natasha13100

I got the following:
spr=400(.12)2/2=2.88
Due to conservation of energy, Uspr=K
K=.8*v2/2
2.88=.4v2 so v=√(7.2)
I use the coordinate where the direction of the normal force is +y and up the incline is +x.
Fy=0 since the block is on the incline.
Fy=N-Gy since N and Gy are in opposite directions.
Gy=-.8*9.8*cos(30)=-98√(3)/25 so N=98√(3)/25
Fx=Ff+Gx since Ff and Gx are in the same direction
Ff=-.2*98√(3)/25=-0.784√(3) and Gx=-.8*9.8*sin(30)=-3.92
therefore, Fx=-0.784√(3)-3.92
F=Fx=ma so a=-(0.784√(3)+3.92)/.8
vf2=0 because the block isn't moving when it reaches its highest point of the incline and vi2=7.2 from earlier. 0=7.2-2((0.784√(3)+3.92)/.8)d so d=3.6/((0.784√(3)+3.92)/.8). sin30=h/d so h=3.6/((0.784√(3)+3.92)/.8)sin30≈0.2728343482
Doesn't the value for h seem a little small?

10. Jul 17, 2013

### TSny

I agree with your final answer. The numerical calculation would be simpler if you wait and substitute numbers into your formula for h:

h=dsinθ=kx2sinθ/(2mg*(μkcosθ+1sinθ))

You don't need to worry about finding the velocity at the bottom of the incline if you set it up as Ef = Ei +Wf, where the initial point is where the block is compressed against the spring and the final point is where the block comes to rest on the incline.

Ef = Ug = mgh
Ei = Usp = kx2/2
Wf = work done by friction = -fkd = -μkmgcosθd = -μkmgcosθh/sinθ