Work done by worker pushing block down frictionless incline at constant speed

[suspenc3]

A 45Kg block slides down a frictionless incline 1.5m long and 0.91m high. A worker pushes up against the block, parallel to the incline, so that the block slides down the incline at a constant speed.

(a)what is the magnitude of the workers force?

(b)How much work is done on the block by the workers force?

(c)The Gravitational force?

(d)The normal force on the block from the surface of the incline?

(e)The Net Force on the block.

m = 45Kg
d = 1.5m
Theta = 37.35
a = 0m/s^2

I drew a F.B.D and came up with the following:

(a)
(-F) - mgsin(theta) = ma = 0
-mgsin(theta) = F
F = -267.5N

(b)
Wa = Fd
Wa = -267.5N(1.5m)
Wa = -401 J

(c)
Wg = -Wa
Wg = 401 J

(d)
N - mgcos(theta) = 0
N = mgcos(theta)
N = 350.6 N

(e)
F(total) = ma
F(total) = 0

 

[whozum]

(a) Correct approach but what is the positive direction?
(d) Careful. Which direction does the normal force act. Does it have any component parallel to the displacement of the box?

 

[suspenc3]

Positive Direction = up the ramp

The normal force is perpendicular to the ramp isn't it?..it doesn't have any component parallel to the displacement i don't think.

So I'm guessing that my answers are wrong?

 

[whozum]

Positive Direction = up the ramp

If positive is up the ramp, then your answer means the guy is pushing down the ramp (you gave a negative answer).

The normal force is perpendicular to the ramp isn't it?..it doesn't have any component parallel to the displacement i don't think.

So if it has no parallel component, did it displace the box? what does that mean about the work done?

 

[suspenc3]

but he must be pushing up the ramp to stop the box from accelerating right?

the normal force wouldn't displace the box i don't think...

Not too sure though..I don't know what to do

 

[whozum]

but he must be pushing up the ramp to stop the box from accelerating right?

Yeah, but if he's pushing -267 ni the UP direction it means he's pushing 267 in the down direction. The mistake you made is in the first line of your work. Redraw your FBD.

the normal force wouldn't displace the box i don't think...

Right it doesn't, if the Force doesn't cause any displacement, it does no work.

 

[suspenc3]

i think i see it...maybe..

I drew the force pushing downwards..when it should have been upwards..soo..

F - mgsin(theta) = ma = 0
F = mgsin(theta)
F = 267.5N

?

 

[whozum]

You goti t.