Block released from spring slides up a ramp

AI Thread Summary
The discussion focuses on solving a physics problem involving a block released from a spring that slides up a ramp. The user initially considers the conservation of energy but struggles with applying the coefficient of friction to the block's motion. They calculate the block's velocity after leaving the spring as 3.553 m/s and explore the components of this velocity along the ramp. Ultimately, they determine the frictional force and gravitational force acting on the block, leading to an acceleration calculation and the distance traveled up the ramp. The user successfully arrives at a solution after seeking clarification on their approach.
Wint
Messages
4
Reaction score
0
[SOLVED] Block released from spring slides up a ramp

Homework Statement


fada240b6cd911d42a59c5a7d7ddd12d.png



Homework Equations


I'm thinking conservation of energy:
K1+U1+WOther=K2+U2
But really I'm not quite sure where to start here.

The Attempt at a Solution


So far, I've calculated the velocity of the block after it leaves the spring, before it hits the ramp and I get that to be 3.553 m/s2. What is tripping me up here is the coefficient of friction, and how to apply that to the block as it travels up the ramp. If I turn the coordinate system so that the x-axis is parallel to the ramp, then we can figure the x component of the velocity to be 3.553*Cos(38) I believe, but I'm not sure if that is useful to me or not.

I think if I could get some confirmation on which direction to take with this that would be very helpful. Thanks!
 
Last edited:
Physics news on Phys.org
Of course I figured it out after asking.

First I figured out that the Fk = n*.200, or cos(38)*(9.8)*(4)*(.200) = 6.178.

Next I figured out the force due to gravity, Fg = sin(38)*(9.8)*(4) = 24.134.

Add the two together and divide by mass (4kg) to get acceleration of 7.578.

Plug that into vf2 = vi2 + 2ad to get d = .833 or so. Sorry to bother everyone!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top