A spring with a spring-constant 2.2 N/cm is compressed 28 cm and released. The 4 kg mass skids down the frictional incline of heigh 43 cm and inclined at a 22 degrees angle

Question

(g = 9.8 m/s^2)
The path is frictionless except for a distance of 0.9 m along the incline which has a coefficient of friction of 0.4

What is the final velocity of the mass?

I decided to solve this into 2 parts.
One part is when it's at the top with the spring which I found was conserving energy
so I had (mgh + .5kx^2) = (mgh + .5mv^2) and found the v when the block released which became the initial v in the incline

Then the incline's equation (wasn't conserved) was E = Efinal - Einitial = .5mvfinal^2 - (.5mvinitial^2 + mgh) = Fx (F being the frictional force)
and I solved for v final

I don't see what I'm doing wrong but I keep getting the wrong answer

Thank you!!

Answer 1

OK, I agree with part1.

1/2 mv^2=1/2 kx^2
v= sqrt(k/m .28^2)=6.57m/s
does that match yours?

Now, work done by friction down the plane= mg*mu*cosTheta*distance
= 4*9.8*.4*cos22*.9= 13.1J

so KE at the base of the plane
1/2 mvf^2=initialGPE+initialKE-friction
= 4*9.8*Vf^2=4*9.8*.43+1/2 4*6.56^2-13.1

=39.2 Vf^2=16.9+86.2-13.1=90.0
Vf=1.52m/s

check all that.

Answer 2

Well so to convert k constant to N/m
would it be 22 N/m instead of 2.2 N/cm?

Answer 3

no, and I erred above, it becomes 220N/m (I put 2200 above).

Answer 4

well since there's an x component of the gravity
wouldn't that affect the force of friction?

For my force i got (0 = theta)
mgsin0 - (mu)mgcos0

is that wrong?

Answer 5

The force of friction is mu*forcenormal

mu*mgCosTheta in this case, theta is 22 deg.

Answer 6

it's still wrong :(

i've tried my way and your way and it's still wrong

Answer 7

okay i got it
thank you so much
i had my signs all wrong
thank you again!

Answer 8

hey mimi what was your final answer?
i have a similar problem and im trying to see if my answer is wrong or not.