Constant velocity and net force

[komender]

Homework Statement

A man is pushing a 75 kg crate at constant
velocity a distance of 12 m across a warehouse.
He is pushing with a force of 225 N at an angle
of 15° down from the horizontal. The coefficient
of friction between the crate and the floor is
0.24. How much work did the man do on the
crate?

Homework Equations

The Attempt at a Solution

Fn = mg + 225sin15 = 794
Fnet = 217.33 - 190.5 = 26.83
26.83*12= 321 J
I did solve the question, but there is one thing that I don't understand:
How can the object have a net force (in this case, 26.83) and still move in constant velocity? Wouldn't it have a constant acceleration instead?

 

[Simon Bridge]

How can the object have a net force (in this case, 26.83) and still move in constant velocity? Wouldn't it have a constant acceleration instead?

A constant net force means a constant acceleration - you are correct. There is no way a net force produces a constant velocity.

 

[komender]

but then the question says that the box is moving at constant velocity when it was pushed 12m...

 

[QED Andrew]

You calculated the net work done on the crate by both the man and frictional force. What is the work done on the crate by the man alone?

In regards to your question concerning the claim of the constant velocity of the crate, either the statement of the question is in error on this point or there is an additional force acting on the crate.

 

[Simon Bridge]

Yeah - what he said: you have to have courage in what you observe.
The problem has described something that cannot happen.

You should check your ideas about force and work - W=Fd ... here, F would be the force exerted by the man in the direction of the motion.

 

[komender]

But aren't you looking at the net force that is generated from the man's push? If I would only account for the force that was exerted by the man, than I would get 225cos15*12 = 2608J, but the correct answer is 320 because I'm taking into account the normal force which is mg + 225sin15 = 794, then I multiply by the coefficient which gives me 190.56N and then I add the two forces (225cos15 - 190.56) which gives me 26.8 N, which I substitute to the formula to get 322 J, which is roughly what they were looking for.
Overall, the question's statement was wrong, and the box couldn't move at constant speed with a net force.

 

[Simon Bridge]

If I would only account for the force that was exerted by the man, than I would get 225cos15*12 = 2608J, but the correct answer is 320 because I'm taking into account the normal force

This is how the thinking went:

The man exerts normal and horizontal components. The total work done by the man is just the sum of the work done by each component.

What is the distance moved in the direction of the horizontal force?
Therefore, how much work is done by the horizontal force?

What is the distance moved in the direction of the normal force?
Therefore, how much work is done by the normal force?

Therefore: what is the total work done by the man?

But you are right - the question does ask for the amount of work done on the crate ... some of the total work is done on the ground (via friction). So the work done on the crate is the work done against friction subtracted from the total work done by the man.

Which would be the net force on the crate, multiplied by the distance traveled.
You could also do it by the work-energy relation and some kinematics - which could be an interesting way to check considering how they think it's a constant velocity.