A 290 kg piano slides 4.6 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40.
(a) Calculate the force exerted by the man.
N
(b) Calculate the work done by the man on the piano.
J
(c) Calculate the work done by the friction force.
J
(d) What is the work done by the force of gravity?
J
(e) What is the net work done on the piano?
J
The force moving the piano down the plane is gravity: mgsinTheta
The retarding forces are friction (mu*mgCosTheta) and the man pushing.
set the retarding force equal to the gravitational force, and you have a.
Work done by mane: his force*distance down the incline
i got 3029.6 for a and still get wrong
of course it is wrong.
forceman+forcefriction=weight down plane
forceman=mgsinTheta-mu*mg*cosTheta
= 290*9.8(sin30-.4*cos30)
Surely you can see that is not 3000N
its incorrect
F = (290*9.8*sin30) - (290*9.8*cos30*0.40) = 436.5 N.
Is this the correct answer?
To find the answers to these questions, we need to understand the concept of work, the forces acting on the piano, and how they affect its motion. Let's break down each part of the problem.
(a) To find the force exerted by the man, we need to consider the forces acting on the piano. In this case, the only horizontal force is the force exerted by the man, which opposes the force of friction. The vertical force of gravity can be broken into two components, one along the incline and the other perpendicular to it. The component along the incline affects the motion of the piano, while the perpendicular component is balanced by the normal force.
The force exerted by the man is equal in magnitude and opposite in direction to the force of friction, which can be calculated using the equation:
Force of friction = coefficient of friction * normal force
The normal force can be found using the vertical component of the force of gravity:
Normal force = mass * gravitational acceleration * cos(angle of inclination)
Substituting the given values:
Normal force = 290 kg * 9.8 m/s^2 * cos(30°)
Normal force = 1450 N * 0.866 (cos 30° is equal to 0.866)
Normal force = 1256.7 N
Now we can calculate the force exerted by the man:
Force of friction = 0.40 * 1256.7 N
Force exerted by the man = -0.40 * 1256.7 N (opposite direction)
Force exerted by the man = -502.7 N (negative value indicates opposite direction)
Therefore, the force exerted by the man is 502.7 N in the direction opposite to the motion of the piano.
(b) The work done by the man on the piano can be calculated using the formula:
Work = force * distance * cos(theta)
Since the force exerted by the man and the displacement are parallel, the angle between them (theta) is 0°, and cos(0°) = 1.
Work done by the man = Force * distance
Work done by the man = 502.7 N * 4.6 m
Work done by the man = 2313.2 J
Therefore, the work done by the man on the piano is 2313.2 Joules.
(c) The work done by the friction force can be calculated using the same formula, but the force and the displacement are in opposite directions, so the angle (theta) is 180°, and cos(180°) = -1.
Work done by the friction force = Force * distance * cos(theta)
Work done by the friction force = -Force of friction * distance (since cos(180°) = -1)
Work done by the friction force = -(0.40 * 1256.7 N) * 4.6 m
Work done by the friction force = -2297.2 J
Therefore, the work done by the friction force on the piano is -2297.2 Joules. The negative sign indicates that work is done against the direction of motion.
(d) The work done by the force of gravity can be calculated using the formula:
Work = force * distance * cos(theta)
The force of gravity is acting vertically downward, while the displacement is along the incline. Thus, the angle between them (theta) is 90°, and cos(90°) = 0.
Work done by the force of gravity = Force of gravity * distance * cos(theta)
Work done by the force of gravity = (mass * gravitational acceleration * sin(theta)) * distance
The component of the gravitational force along the incline is given by:
Force of gravity along incline = mass * gravitational acceleration * sin(angle of inclination)
Work done by the force of gravity = (290 kg * 9.8 m/s^2 * sin(30°)) * 4.6 m
Work done by the force of gravity = 578.3 J
Therefore, the work done by the force of gravity on the piano is 578.3 Joules.
(e) The net work done on the piano is the sum of the work done by various forces. Since forces can be positive or negative depending on their direction, we can calculate the net work by taking the algebraic sum of the individual works.
Net work done on the piano = Work done by the man + Work done by the friction force + Work done by the force of gravity
Net work done on the piano = 2313.2 J + (-2297.2 J) + 578.3 J
Net work done on the piano = 594.3 J
Therefore, the net work done on the piano is 594.3 Joules.