Work/Energy Problem: 50g Cube on 30 Degree Slope

  • Thread starter Thread starter holtvg
  • Start date Start date
AI Thread Summary
A 50 g plastic cube slides on a 30-degree slope, compressed against a spring with a constant of 25 N/m. When released, the cube's motion is affected by kinetic friction (coefficient of 0.20) and gravitational forces. The discussion revolves around calculating the distance the cube travels up the slope before reversing direction, considering potential and kinetic energy, as well as work done against friction. Participants express confusion about incorporating kinetic energy and the effects of friction in their calculations. The consensus suggests that the distance will be less than the 50 cm calculated without friction, indicating the need for a more precise approach to account for all forces involved.
holtvg
Messages
17
Reaction score
0

Homework Statement



A 50 g plastic cube slides up and down a 30 degree slope with friction. The plastic cube is pressed against a spring at the bottom of the slope, compressing the spring 10 cm. The spring constant is 25 N/m. When the plastic cube is released, what total distance will it travel up the slope before reversing direction if the coefficient of kinetic friction is 0.20. How far will the plastic cube travel up the slope?

Homework Equations



spe=1/2kx^2 gpe=mgh ke=1/2mv^2 wf=uNd

The Attempt at a Solution



spe=wf+gpe

1/2*25*.1^2=-.20*9.8*cos(30 degrees)*d+.05*9.8*h

solve for d, can't solve equation, two unknowns? Tried to figure out h but couldn't.
 
Last edited:
Physics news on Phys.org
h is a vertical distance. So it makes a right angled triangle.

the hypotenuse is d and the height is h, and the angle is 30. I think you can find h in terms of d now.
 
You can put h = d*sinθ.
In addition to the frictional force, one more force in acting on the plastic block. Which one?
 
There's the x component of gravity and the normal force, but I'm not sure the x component of gravity does work. I know with respect to the cube the y component of gravity gives it potential energy. Does the spring do work tho, as it does apply a force f=-kx over a distance of 10 cm, that energy then becomes kinetic energy, then is transferred to friction and potential. Should i have kinetic energy in the equation? Or the moment just after the spring hits the cube.

ke=1/2mv^2
 
When the cube is moving up, the frictional force and the x-component of the gravitational force act in the opposite direction. If d is the distance covered by the cube along the slope before coming to rest, then KE = (fg + fr )*d. You need consider PE separately.
 
Ok you can't set h=dsin(theta) that gives you the wrong answer as i get about 140 cm. I did the math for the cube without friction and it moves about 50 cm, so if there was friction it would obviously be less than 50 cm.
 
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