Question regarding "work" in physics?

[The Wolf]

v=0.6m/s, F=35N, d=24m
W = Fd = 35*24 = 840J
Power = P = Fv = 35*0.6 = 21J/s
.,.,.

 

[imported_high-lighter]

Since the speed is constant, that means the box does not accelerate, thus no additional force is applied to increase speed
=> F=ma, when a=0, then F=0, thus, speed remains constant(because v = u + at and v = u since a = 0)
So, the guy use only 35N to push the box, which is just enough to overcome the 35N frictional force, and once the frictional force is overcome, the box travels at its original speed, u.
Overally, F - friction force = ma. Since ma=0,
=> F - friction force = 0
<=> F = friction force
<=> F = 35N

1) Work done, W=Fd
<=> W = 35 x 24
<=> W = 840J or W = 840Nm

2) Power(J/s), P = Fv
where v can be broken down into d/t
<=> P = Fd/t
<=> P = Fd x (1/t)
You can see P is actually equal to W.Done/time
However, here I use the equation, P = Fv
<=> P = 35 x 0.6
<=> P = 21w (watt) or P = 21J/s

 

[imported_high-lighter]

Since the speed is constant, that means the box does not accelerate, thus no additional force is applied to increase speed
=> F=ma, when a=0, then F=0, thus, speed remains constant(because v = u + at and v = u since a = 0)
So, the guy use only 35N to push the box, which is just enough to overcome the 35N frictional force, and once the frictional force is overcome, the box travels at its original speed, u.
Overally, F - friction force = ma. Since ma=0,
=> F - friction force = 0
<=> F = friction force
<=> F = 35N

1) Work done, W=Fd
<=> W = 35 x 24
<=> W = 840J or W = 840Nm

2) Power(J/s), P = Fv
where v can be broken down into d/t
<=> P = Fd/t
<=> P = Fd x (1/t)
You can see P is actually equal to W.Done/time
However, here I use the equation, P = Fv
<=> P = 35 x 0.6
<=> P = 21w (watt) or P = 21J/s

 

[Leonski]

What is the work done when a person pushes a heavy box along a rough floor at a constant speed of 0.60m/s, against a frictional force of 35N for a distance of 24m?

How many Joules per second is delivered by the person?

Workings appreciated. Thanks. 10 points best answer.

 

[high-lighter]

Since the speed is constant, that means the box does not accelerate, thus no additional force is applied to increase speed
=> F=ma, when a=0, then F=0, thus, speed remains constant(because v = u + at and v = u since a = 0)
So, the guy use only 35N to push the box, which is just enough to overcome the 35N frictional force, and once the frictional force is overcome, the box travels at its original speed, u.
Overally, F - friction force = ma. Since ma=0,
=> F - friction force = 0
<=> F = friction force
<=> F = 35N

1) Work done, W=Fd
<=> W = 35 x 24
<=> W = 840J or W = 840Nm

2) Power(J/s), P = Fv
where v can be broken down into d/t
<=> P = Fd/t
<=> P = Fd x (1/t)
You can see P is actually equal to W.Done/time
However, here I use the equation, P = Fv
<=> P = 35 x 0.6
<=> P = 21w (watt) or P = 21J/s