Work/Energy problem at an angle

  • Thread starter Thread starter Heather_
  • Start date Start date
  • Tags Tags
    Angle
AI Thread Summary
A block of ice slides down a frictionless ramp at a 50° angle while being pulled by a 60 N force. The kinetic energy of the block increases by 85 J over a distance of 0.65 m. The initial calculation for gravitational force in the x-direction was incorrect due to the unnecessary use of the cosine factor. Correcting this mistake reveals that the work done would have resulted in an additional 25.6 J of kinetic energy if the rope had not been attached. The discussion highlights the importance of accurately applying trigonometric functions in physics problems.
Heather_
Messages
4
Reaction score
0
1. Homework Statement

In the figure below, a block of ice slides down a frictionless ramp at angle θ = 50°, while an ice worker pulls up the ramp (via a rope) with a force of magnitude Fr = 60 N. As the block slides through distance d = 0.65 m along the ramp, its kinetic energy increases by 85 J. How much greater would its kinetic energy have been if the rope had not been attached to the block?

2. Homework Equations
W=ΔEk
W=F*d*cosθ

3. The Attempt at a Solution

First, I solved for the gravitational force in the X direction: (Fgx-60N)*.65m*cos50=85J
Fgx = 263.44 N
Then, I plugged it back into the work equation to see how much work would be done then: 263.44N*.65m*cos50=W
W=110.06J=ΔEk
110.6J-85J=25.6J
Yet, this is not the right answer. What am I doing wrong?
 
Physics news on Phys.org
Heather_ said:
First, I solved for the gravitational force in the X direction: (Fgx-60N)*.65m*cos50=85J
The x-component of gravity and the 60 N force are already parallel to the ramp, so no need for the cos50 factor.
 
Wow. That was a silly mistake. Thank you! :)
 
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