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

[revere21]

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

PE_G = 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.

 

[revere21]

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.

 

[Tong32]

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

PE_G = 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).

 

[Tong32]

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.

 

[rwisz]

I would like to commend your attempt at solving this problem and your willingness to seek clarification. Your line of thinking is correct in regards to the work done by friction being negative. This is because the friction force is opposing the motion of the block, meaning it is acting in the opposite direction of the displacement. In physics, work is defined as the product of the force and the displacement in the direction of the force. Therefore, when the displacement and force are in opposite directions, the work done is negative.

For the work done by the string, we can use the same equation as before, W = F(parallel)*d, but now we need to find the parallel component of the force. The string is parallel to the incline, so the angle between the force and displacement is 30 degrees. Therefore, the parallel component of the force is F*cos30 = F*0.87. Now we can calculate the work done by the string as (F*0.87)*(4 m) = 3.48 F. This value will be greater than the work done by friction as the string is doing work to pull the block up the incline.

To find the value of F, we can use the fact that the block is moving at constant speed, meaning the net force in the direction of motion is zero. This means that the force of gravity, the force of friction, and the force of the string must all balance each other out. Therefore, F = mg + 7 N = (1.6 kg)(9.8 m/s^2) + 7 N = 22.88 N.

Plugging this value of F into our equation for the work done by the string, we get (22.88 N*0.87)*(4 m) = 80.08 J. Therefore, the work done by the string is 80.08 J.

I hope this explanation helps and allows you to better understand the concept of work and how to properly calculate it in various scenarios. Keep up the good work!