Work Pulley Problem with constant speed

[Ineedhelpwithphysics]

Homework Statement

Look at picture provided

Relevant Equations

W = F * delta x

For A the 1.2 kg block is being pulled by gravity hence work is done downwards which will make work positive since it's going with the same direction as the force.

1.2 * 9.8 = 11.76 N pulled downwards
Work = F*d
11.76*0.75 = 8.82 J

The tension is the other force and since the thing is going at constant speed:

11.76-T = 0

T = 11.76
but the work for the tension will be negative due to the fact it's opposite to the motion so

-11.76*0.75 = -8.82J

For b: the forces that are acting on the box is tension and friction, since the thing is moving at constant velocity T = Fn
2kg*9.8 = 19.6
T- Fn = 0
11.76 = Fn

The work due to the tension on the 2kg is:
11.76 * 0.75 = 8.82J

Work due to friction on 2kg box is:
-11.76 * 0.75 = -8.82J

For C) I approached this two ways
since the system is moving at constant speed

Wnet = 1/2 m (vf^2 - vi^2)

Wnet = 0 due to the fact its constant speed

The second approach i did was just adding up all the work i calculated.

Can anyone clarify to me if constant speed always means 0 work.

 

[haruspex]

Homework Statement: Look at picture provided
Relevant Equations: W = F * delta x

For A the 1.2 kg block is being pulled by gravity hence work is done downwards which will make work positive since it's going with the same direction as the force.

1.2 * 9.8 = 11.76 N pulled downwards
Work = F*d
11.76*0.75 = 8.82 J

The tension is the other force and since the thing is going at constant speed:

11.76-T = 0

T = 11.76
but the work for the tension will be negative due to the fact it's opposite to the motion so

-11.76*0.75 = -8.82J

For b: the forces that are acting on the box is tension and friction, since the thing is moving at constant velocity T = Fn
2kg*9.8 = 19.6
T- Fn = 0
11.76 = Fn

The work due to the tension on the 2kg is:
11.76 * 0.75 = 8.82J

Work due to friction on 2kg box is:
-11.76 * 0.75 = -8.82J

For C) I approached this two ways
since the system is moving at constant speed

Wnet = 1/2 m (vf^2 - vi^2)

Wnet = 0 due to the fact its constant speed

The second approach i did was just adding up all the work i calculated.

Can anyone clarify to me if constant speed always means 0 work.



View attachment 340739

All good.
Yes, the net work done on a body is always its gain in KE.

 

[kuruman]

Can anyone clarify to me if constant speed always means 0 work.

Constant speed always means zero net work, i.e. the sum of all the works done of the body is zero. It does not mean that no work whatsoever is done on the body.