Block, spring, and incline- work/energy

[natasha13100]

Homework Statement

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 μ_k=.2. (See the attached picture.) Find the maximum vertical height h reached by the block.

Homework Equations

U_spr=potential energy of a spring=kx²/2 where k is the spring constant
K=kinetic energy=1/2mv²
U_g=potential energy due to gravity=mgh
G=gravitational force=mg where g=force due to gravity=-9.8m/s²
F_f=force due to friction (kinetic)=μ_kN
F_tot=total force=ƩF
F=force=ma
v²=v_i²+2ax where v=velocity, a=acceleration, and v_i=initial velocity
other equations may be used

The Attempt at a Solution

I need help understanding where I went wrong and how the problem should be done.
U_spr=kx²/2=(400)(.12)²/2=2.88
K=U_spr
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=-N_y=-Ncos(θ)
N=-G/cos(θ)=7.84/cos(30°)
F_f=-μ_kN=-.2*7.84/cos(30°)=1.568/cos(30°)
F_toty=ƩF_y=7.84/cos(30°)-7.84-1.568/cos(30°)=-.5977
v_0y=3√(2)cos(30°)/5=3√(6)/10
F_toty=ma_y
a_y=F_toty/m=-.5977/.8=-.747125
v_y²=v_0y²+2a_yh
v_y²=0 when the block is at its maximum height
0=(3√(6)/10)²+2(-.747125)h
1.49425h=5.4
h≈3.613853104

 

[TSny]

Welcome to PF!

Homework Equations

U_spr=potential energy of a spring=-kx where k is the spring constant

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

I assume right is in the +x direction and up is in the +y direction.

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.

 

[natasha13100]

Ok so I'm still doing something wrong. Here are some of my latest thoughts on the problem:
U_spr=kx²/2
Due to conservation of energy, U_spr=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.
F_y=0 since the block is on the incline.
F_y=N-G_y since N and G_y are in opposite directions.
G_y=mg/cosθ so N=mg/cosθ
F_x=F_f+G_x since F_f and G_xare in the same direction
F_f=μ_kN=μ_kmg/cosθ and G_x=mg/sinθ
therefore, F_x=μ_kmg/cosθ+mg/sinθ.
I've tried multiple approaches here:

1.F=F_x=ma so a=F_x/m=μ_kg/cosθ+g/sinθ
v_f²=0 because the block isn't moving when it reaches its highest point of the incline and v_i²=kx²/m from earlier. v_f²=v_i²+2ad so 0=kx²/m+2(μ_kg/cosθ+g/sinθ)d and d=kx²/(2mg*(μ_k/cosθ+1/sinθ)). sinθ=h/d so h=dsinθ=kx²sinθ/(2mg*(μ_k/cosθ+1/sinθ)).

2. w=Fcosθx so w=-F_xd=-mgd(μ_k/cosθ+1/sinθ)
E_i=E_f+W
K=U_g+W
K=U_spr=kx²/2, U_g=mgh
kx²/2=mgh-mgd(μ_k/cosθ+1/sinθ)
*since d=h/sinθ, kx²/2=mgh-mgh(μ_k/cosθ+1/sinθ)/sinθ
therefore, h=kx²sinθ/(2mg(μ_ktanθ+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.

 

[TSny]

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

Your method 1 looks good except for the expressions for G_x and G_y.

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 U_grav. 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 E_i=E_f+W. Can you spot it?]

 

[TSny]

I use the coordinate where the direction of the normal force is +y and up the incline is +x.

On the slope, is F_x 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?

 

[natasha13100]

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

 

[TSny]

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

Yes. That looks good.

 

[natasha13100]

Thank you so much!
So that means N=-mgcosθ
F_x=F_f+G_x
F_f=μkN=-μkmgcosθ and G_x=-mgsinθ
therefore, F_x=-μkmgcosθ-mgsinθ.
F=F_x=ma so a=F_x/m=-μkgcosθ-gsinθ
v_f²=0
v_f²=v_i²+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)).

 

[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.
F_y=0 since the block is on the incline.
F_y=N-G_y since N and G_y are in opposite directions.
G_y=-.8*9.8*cos(30)=-98√(3)/25 so N=98√(3)/25
F_x=F_f+G_x since F_f and G_x are in the same direction
F_f=-.2*98√(3)/25=-0.784√(3) and G_x=-.8*9.8*sin(30)=-3.92
therefore, F_x=-0.784√(3)-3.92
F=F_x=ma so a=-(0.784√(3)+3.92)/.8
v_f²=0 because the block isn't moving when it reaches its highest point of the incline and v_i²=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?

 

[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 E_f = E_i +W_f, 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.

E_f = U_g = mgh
E_i = U_sp = kx²/2
W_f = work done by friction = -f_kd = -μ_kmgcosθd = -μ_kmgcosθh/sinθ