Calculating Forces and Work Done on a Piano Sliding Down an Incline

[confusedaboutphysics]

A 300 kg piano slides 4.6 m down a 30° 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.
(a) Calculate the force exerted by the man.

i got F = mu(k)mgcos(angle) + mgsin(angle) = 0
so F = .40(300)(9.8)cos(30) + 300(9.8)sin(30) = 451.55N

(b) Calculate the work done by the man on the piano.
(c) Calculate the work done by the friction force.
(d) What is the work done by the force of gravity?
(e) What is the net work done on the piano?

ok so I'm having problems with parts b and c. I thought I could use the equation W = FD for part b. so W = 451.55(4.6) = 2077.13 J. is 4.6m d? help please!

 

[Doc Al]

(a) Calculate the force exerted by the man.
i got F = mu(k)mgcos(angle) + mgsin(angle) = 0
so F = .40(300)(9.8)cos(30) + 300(9.8)sin(30) = 451.55N

Careful. What are the directions of the forces?

Edit: Your answer is correct, but your equation is not. You must have copied it wrong.

 

[pizzasky]

Hi!

Now, the piano is moving downward along the incline but the force exerted by the man is upward along the incline. Hence, work done by the man is negative.

You are right in saying that d= 4.6m, as the line of action of the man's force and the motion of the piano are parallel, but you'll need to add a negative sign in front of your answer for part (b).

The same goes for part (c), as frictional force is opposite to the piano's motion!

Hope this helps!

 

[confusedaboutphysics]

thanks guys! i understand b now..but I'm still stuck on c. is this the right equation for part c? Wfr = Ffr (dcos30) = .40(300)(-9.8) (4.6cos30) = 4684.85 J?

 

[lightgrav]

How far is the contact-point of the F_fr displaced?
Is this F_fr parallel, anti-parallel, or at 30 deg to dx?

isn't the surface Normal Force = m g cos30 ?

 

[pizzasky]

Hi!

Your final answer seems correct, but why do you have the expression "(4.6cos30)"?

This seems to imply that you are taking a component of the distance traveled by the piano, which does not seem right, as the direction of motion of the piano is parallel (though in opposite direction to) the frictional force.

Instead, the cos30 is meant for taking the component of the piano's weight which is perpendicular to the incline, to be used later in calculating the frictional force.

Hope you get what I mean...

 

[Doc Al]

..but I'm still stuck on c. is this the right equation for part c? Wfr = Ffr (dcos30) = .40(300)(-9.8) (4.6cos30) = 4684.85 J?

While your answer is correct (except for the sign), the way you've grouped your factors doesn't make sense. The work done by any force is just force times the displacement parallel to that force. The friction force (which you calculated to solve part a) is \mu m g \cos \theta acting up the incline; the displacement is given as 4.6 m down the incline.