Question
A 290 kg piano slides 4.0 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). The effective coefficient of kinetic friction is 0.40
Answers
(a) Calculate the force exerted by the man.
(290*9.8*sin 30deg)-(290*9.8cos30deg*0.40)=436.5N
(b) Calculate the work done by the man on the piano.
284.2*(-0.40)=-1136.8J
(c) Calculate the work done by the friction force.
-3937.6J
(d) What is the work done by the force of gravity?
5684 J
(e) What is the net work done on the piano?
0J (ZERO J)
(290*9.8*sin 30deg)-(290*9.8cos30deg*0.40)=436.5N
(b) Calculate the work done by the man on the piano.
284.2*(-0.40)=-1136.8J
(c) Calculate the work done by the friction force.
-3937.6J
(d) What is the work done by the force of gravity?
5684 J
(e) What is the net work done on the piano?
0J (ZERO J)
4 sin 30 = 2 meters down
change in potential energy = work done by gravity = m g h = 290*9.8*2 Joules
work done by man = -F (4)
work done by friction = - m g cos 30 (.4)(4)
since it does not accelerate, the total net work done on it is zero
- 4 F -290*9.8 (cos 30) (1.6) + 290*9.8*2 = 0
so
4 F = 290(9.8) [ 2 - 1.6 cos 30 ]
change in potential energy = work done by gravity = m g h = 290*9.8*2 Joules
work done by man = -F (4)
work done by friction = - m g cos 30 (.4)(4)
since it does not accelerate, the total net work done on it is zero
- 4 F -290*9.8 (cos 30) (1.6) + 290*9.8*2 = 0
so
4 F = 290(9.8) [ 2 - 1.6 cos 30 ]
µk x (m x g) x d x cos 30º
0.40 x (290 x 9.81) x 4.0 x cos 30º
hopefully that works
0.40 x (290 x 9.81) x 4.0 x cos 30º
hopefully that works
that is the magnitude of the work done by friction
You did the whole problem with forces, I did it with energy. Same results
You did the whole problem with forces, I did it with energy. Same results
( I did the problem before md8 posted the second time listing the specific questions.)