A 6.0 kg box is sliding up a 8.0 m long friction incline at a constant speed at an angle of 30 degrees by a force 50 N parallel to the incline. The coefficient of kinetic friction is 0.1. The work done against friction is??
2007-03-11 06:04:56 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Ok, here you have a Force of 50 Newtons parallel to the incline.
Then you can calculate the force of friction. Considering g = 10 m /s^2 >> gravity
the weight of the box = 60 N
Decomposing the force, we can find the normal force that will be : 60*cos(30)
Force of friction = 60*cos(30)*0.1 = 6*sqrt(3) / 2 = 3*sqrt(3) N
The work done by the force of friction = 24*sqrt(3) Joules
The work done by the force of 50 N = 50*8 = 400 Joules
Hope that might help you
2007-03-11 06:31:55 · answer #1 · answered by anakin_louix 6 · 0⤊ 0⤋
Well, work is equal to Fs*cos(theta) and we know s, the displacement as well as the angle, theta. I'm pretty sure the force of an object on an incline plane is equal to mg*sin(theta) so you can calculate the force. Frictional force is equal to the normal force times the coefficient of friction right? I would subtract the frictional force from the applied force to find the net force. Then, multiply this Force times the distance times cosine of 30* to find the work done.
2007-03-11 06:36:05 · answer #2 · answered by sprintdawg007 3 · 0⤊ 0⤋