Box dragged along resistant surface, work = ?

Click For Summary
SUMMARY

The discussion focuses on calculating the work done on a box being dragged over a rough surface by a force F at an angle θ. The user initially attempts to derive the work done using the equation W = [F cos θ - μ(mg - F sin θ)] d but realizes this is incorrect. The correct approach involves recognizing that the work done by friction must be calculated separately, using the relation W = F · Δx, where friction is treated as a nonconservative force. The user successfully solves for the work done by friction and the final speed of the box but struggles to finalize the work done on the block.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of work and energy
  • Knowledge of frictional forces and coefficients
  • Basic proficiency in vector mathematics
NEXT STEPS
  • Study the principles of work-energy theorem in physics
  • Learn about calculating work done by nonconservative forces
  • Explore one-dimensional motion equations for final velocity calculations
  • Investigate the effects of varying friction coefficients on work done
USEFUL FOR

Students and educators in physics, mechanical engineers, and anyone interested in understanding the dynamics of objects in motion on surfaces with friction.

lizzyb
Messages
167
Reaction score
0
box dragged along resistant surface, work == ??

1) We have a box with mass m being dragged over a rough horizontal surface by a constant force F at an angle @ above the horizontal; it travels d meters and there is a kinetic frictional coefficient of u. What is the work done on the block?

2) What is the work done by the force of friction.

3) Determine it's final speed assuming it's initial velocity was 0.

Attempted Answer:

1)
Fx = F cos @ - uN = m a_x (1)
Fy = F sin @ + N - m g = 0 (2)

From (2) we have N = m g - F sin @, substituting into (1) we have

F cos @ - u (m g - F sin @) = m a_x

Therefore, W = [ F cos @ - u (m g - F sin @)] d

But this is the wrong answer??

2) I don't know how to set this up; what are we solving for?

3) If I can come up with the right answer for (1), I can determine the acceleration and determine it's final speed via one of the one-dimensional motion equations.

please help! :-)
 
Physics news on Phys.org
Think about it this way: how much work must be done on the block for it to move d meters?

Pretend that there isn't friction. The amount of work done on the block is now [tex]\vec{F} \cdot \Delta \vec{x}[/tex], or F*d*cos@.

However, there is friction. So the amount of work done on the block is greater than it would have been if there were no friction.

Look at your expression:
W = [ F cos @ - u (m g - F sin @)] d

This means that it takes less work to move the block if there is friction, which does not make much sense.
2) I don't know how to set this up; what are we solving for?
You are solving for the work done by friction. Use the relation that [tex]W = \vec{F} \cdot \Delta \vec{x}[/tex] to calculate the work done by friction. Don't forget that friction is a nonconservative force.

3) If I can come up with the right answer for (1), I can determine the acceleration and determine it's final speed via one of the one-dimensional motion equations.
You don't need to answer (1) to get the answer for (3). You can already find the acceleration.
 
Thank you for your reply and aid. I solved (2) and (3), but have yet to correctly answer (1). I tried W = [ F cos @ + u (m g - F sin @)] d but to no avail. Any ideas?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
55
Views
6K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K