Work done by string, friction, pulling mass up an inclined plane

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a block being pulled up a rough incline at a constant speed. The problem includes calculations related to gravitational potential energy, work done by friction, and work done by the string, with specific values and conditions provided.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of gravitational potential energy and work done by friction, with some questioning the sign of the work done by friction due to its direction relative to motion.
  • There are attempts to understand the work done by the string, with participants expressing uncertainty about how to incorporate various forces and distances into their calculations.
  • Conceptual questions are raised regarding the work done by the normal force and the relationship between potential and kinetic energy when the block is released.

Discussion Status

Participants are actively engaging with the problem, sharing their reasoning and calculations. Some guidance has been offered regarding the nature of work done by friction and the conditions under which it is considered negative. Multiple interpretations of the work done by the string and the implications of energy conservation are being explored.

Contextual Notes

The problem is constrained by specific values for mass, incline angle, and friction force, which are central to the calculations being discussed. There is also an emphasis on understanding the implications of energy transfer and the role of different forces acting on the block.

revere21
Messages
10
Reaction score
0

Homework Statement


A block of mass 1.60 kg is pulled up a rough 30 degree incline at constant speed by a string parallel to the surface. The friction force on the block is 7.00 N. The incline is 4.00 m long and 2.00 m high.

(1) How much gravitational potential energy did the block gain?
(2) How much work was done by friction?
(3) How much work was done by the string?


Homework Equations


PEG = m*g*y
W = F(parallel)*d


The Attempt at a Solution


(1) [using equation 1] The gravitational potential energy gained was just (1.6 kg)(9.8 m/s2)(2 m) = 31.4 kg*m2/s2, which is 31.4 J.

(2) [using equation 2] The work done by friction was (7 N)(4 m) = 28 J. BUT, the answer key says -28 J, so my best reasoning for why is that, the friction force of 7 N is opposite the direction of motion, which makes this value negative. Can someone please tell me if my line of thinking is correct, and add an explanation that will allow me to (in the future) understand when the value should be negative and when it should not?

(3) For the amount of work done by the string, I really have no idea. It would need to be greater than the 28 J of work done by friction in order to pull the object up the incline (forgive me if this is a gross oversimplification), and it would also be combatting gravity. And surely the mass must be figured in as well. And, I believe, the distance traveled. Can anyone help me in setting up an equation that would correctly integrate everything that must be included and lead me in the right direction?

Thank you all in advance.
 
Physics news on Phys.org
In addition, two conceptual questions are asked of the problem.

(1) Why is the work done on the block by the normal force zero?
I believe this would be because the normal force of the block is equal to the weight, but I am not sure.

(2) Imagine the string is now released. The block starts from rest and slides down the rough plane. At the bottom, its kinetic energy is:
I put that it is equal to its potential energy at the top. My reasoning is due to conservation of energy.

Thank you all again.
 
revere21 said:

Homework Statement


A block of mass 1.60 kg is pulled up a rough 30 degree incline at constant speed by a string parallel to the surface. The friction force on the block is 7.00 N. The incline is 4.00 m long and 2.00 m high.

(1) How much gravitational potential energy did the block gain?
(2) How much work was done by friction?
(3) How much work was done by the string?


Homework Equations


PEG = m*g*y
W = F(parallel)*d


The Attempt at a Solution


(1) [using equation 1] The gravitational potential energy gained was just (1.6 kg)(9.8 m/s2)(2 m) = 31.4 kg*m2/s2, which is 31.4 J.

(2) [using equation 2] The work done by friction was (7 N)(4 m) = 28 J. BUT, the answer key says -28 J, so my best reasoning for why is that, the friction force of 7 N is opposite the direction of motion, which makes this value negative. Can someone please tell me if my line of thinking is correct, and add an explanation that will allow me to (in the future) understand when the value should be negative and when it should not?

(3) For the amount of work done by the string, I really have no idea. It would need to be greater than the 28 J of work done by friction in order to pull the object up the incline (forgive me if this is a gross oversimplification), and it would also be combatting gravity. And surely the mass must be figured in as well. And, I believe, the distance traveled. Can anyone help me in setting up an equation that would correctly integrate everything that must be included and lead me in the right direction?

Thank you all in advance.

(2) The work done by friction on the block is intrinsically negative.
(3) W = F x d W = (mgsinθ + Ff) x d (1.6kg(sin30) + 7.0N) x 4.0 m = 59.4
Since the block is moving at a constant speed we assume the acceleration is zero. This then allows us to know that the force normal is equal to the opposing forces (friction and the force of gravity in the x direction).
 
revere21 said:
In addition, two conceptual questions are asked of the problem.

(1) Why is the work done on the block by the normal force zero?
I believe this would be because the normal force of the block is equal to the weight, but I am not sure.

(2) Imagine the string is now released. The block starts from rest and slides down the rough plane. At the bottom, its kinetic energy is:
I put that it is equal to its potential energy at the top. My reasoning is due to conservation of energy.

Thank you all again.

(1) Because it is perpendicular to the direction of motion
(2) The kinetic energy is less than its potential energy at the top because some of the energy is transferred into heat due to friction.
 

Similar threads

Replies
2
Views
679
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
8K
  • · Replies 33 ·
2
Replies
33
Views
4K
  • · Replies 27 ·
Replies
27
Views
10K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
3K
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K