Calculating the Work Done by Normal Force on a Sliding Piano

[tica86]

A 330-kg piano slides 3.6 m down a 28º incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40

what is the work done by the normal force?

If someone could let know how to find FN,

would it be mgcos28??

 

[venkatg]

Workdone is F * distance.

Normal force as you said is mgcos28. But the distance is zero!
So no work is done by the normal force.

 

[fizzynoob]

your Fn is correct

 

[tica86]

Workdone is F * distance.

Normal force as you said is mgcos28. But the distance is zero!
So no work is done by the normal force.

Ok, I understand the definition of Work done but how do you know that the distance is zero?

Since the piano is sliding in the horizontal direction the 3.6 distance is horizontal and since normal force is vertical there is no distance??

If there was acceleration the net work done on the piano would NOT be 0 right?

 

[venkatg]

Ok, I understand the definition of Work done but how do you know that the distance is zero?

Since the piano is sliding in the horizontal direction the 3.6 distance is horizontal and since normal force is vertical there is no distance??

If there was acceleration the net work done on the piano would NOT be 0 right?

Yes the the normal force does not cause any movement and so distance is zero.
This is the case even if there was acceleration in the horizontal direction (in this case along the incline)

 

[tica86]

Yes the the normal force does not cause any movement and so distance is zero.
This is the case even if there was acceleration in the horizontal direction (in this case along the incline)

Ok,thanks.