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
(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)
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 ]
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
that is the magnitude of the work done by friction
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.)
To solve this problem, we need to consider the forces acting on the piano.
First, let's determine the gravitational force acting on the piano. The gravitational force can be calculated using the formula:
F_gravity = m * g
Where:
m = mass of the piano = 290 kg
g = acceleration due to gravity = 9.8 m/s^2
F_gravity = 290 kg * 9.8 m/s^2
F_gravity = 2842 N
Next, we need to calculate the component of the gravitational force acting parallel to the incline. This can be done using the formula:
F_parallel = F_gravity * sin(θ)
Where:
θ = angle of the incline = 30°
F_parallel = 2842 N * sin(30°)
F_parallel = 1421 N
This parallel force is acting down the incline and is opposed by the force of kinetic friction and the force applied by the man.
The force of kinetic friction can be calculated using the formula:
F_friction = μk * N
Where:
μk = coefficient of kinetic friction = 0.40
N = normal force
The normal force is the force perpendicular to the incline and can be calculated using the formula:
N = m * g * cos(θ)
N = 290 kg * 9.8 m/s^2 * cos(30°)
N = 2507 N
Now, we can calculate the force of kinetic friction:
F_friction = 0.40 * 2507 N
F_friction = 1003 N
Since the man is pushing back on the piano parallel to the incline, we can assume that the force he applies is equal in magnitude and opposite in direction to the force of kinetic friction. Therefore, the force applied by the man is also 1003 N.
Now, we can calculate the net force acting on the piano parallel to the incline:
Net Force = F_applied - F_friction
Net Force = 1003 N - 1003 N
Net Force = 0 N
The net force is zero, which means that the piano is not accelerating. The man is applying a force equal in magnitude to the force of kinetic friction, thus balancing out the forces and keeping the piano from accelerating.
In summary, in order to solve this problem, we needed to calculate the gravitational force, the component of the gravitational force parallel to the incline, the force of kinetic friction, and the force applied by the man. By balancing the forces, we found that the net force parallel to the incline is zero, meaning the piano remains at a constant velocity.