network done on the box?

Question

A 6.0 kg box slides up a 8.0 m long friction incline at an angle of 30 degrees by a force 50 N parallel to the incline. The velocity of the box is constant throughout and the coefficient of kinetic friction is 0.1. The network done on the box is??

a) -03.00
b) 96.55
c) -60.30
d) 305.00
e) 00.00

Answer 1

Puzzling problem. At first I thought more information was provided just to confuse things. Well it does. It doesn't seem the data is compatible.

The 50 N force is applied for the 8 m trip up the incline. It's directed parallel to the incline so the work it does is
F*d = 50 N*8 m = 400 Nm

The friction is
Ff = mu*N
where N is the force normal to the incline
N = 6 kg*g*cos30
N = 6 kg*9.8 m/s^2*cos30 = 50.9 N

So
Ff = mu*N = 0.1*50.9 N = 5.09 Newtons

The work that was done on the friction is
W = Ff*d = 5.09 N*8 m = 40.7 Nm

The force did 400 Nm of work (my 2nd paragraph). That's 400 Joules of energy that went somewhere. After the work done on the friction, that leaves
400 Nm - 40.7 Nm = 359.3 Nm
so it seems that this remainder is the net work done on the box. But that's not one of the choices.

Looking at potential energy gain of the box:
PE = m*g*h
PE = 6 kg*9.8 m/s^2*sin30*8 m = 235.2 Joules

So the block gained 235.2 Joules of potential energy. The velocity stayed the same it says, so the KE didn't change.

It should be true that the work done on the box is the gain in potential energy, or heat, or a combination of such things. This problem does not compute.

I've re-read this several times looking for another way to look at it. I think it's a bad question. I do have a pretty good record with Physics questions here.

Answer 2

i have no idea what u r talking about

Answer 3

no