A father pushes horizontally on his daughter's sled to move it up a snowy incline, as illustrated in the figure, with h = 3.9 m and θ = 11°. The total mass of the sled and the girl is 35 kg and the coefficient of kinetic friction between the sled runners and the snow is 0.20. If the sled moves up the hill with a constant velocity, how much work is done by the father in moving it from the bottom to the top of the hill?
Force component exerted parallel to slope by man is Fp
Fp = (0.2)(35x9.81xcos11) + (35x9.81xsin11)
(note Fp is up hill, both friction and weight forces are down hill)
Fp = 133 Newtons parallel to slope
motion is parallel to slope
so
work = Fp * hypotenuse length
= 133 * 3.9/sin 11 = 2717 Joules
(force exerted by father)+(friction)-weight along the slope = 0
F+(0.2)(35x9.81xcos11)-(35x9.81xsin11)= 0
F= -1.89N (hey, why it is negative? It is wrong. I need more information)
work done = F.S
oh, i know why i am wrong.it is because i take friction a wrong direction. thanks very much!(Damon is correct!)
shouldnt the work = Fp*d*cos(angle between) ?
e=mc^2
To find the work done by the father in moving the sled from the bottom to the top of the hill, we can use the work-energy principle. The principle states that the work done on an object is equal to the change in its kinetic energy.
In this case, since the sled is moving up the hill with a constant velocity, its kinetic energy does not change. Therefore, the work done by the father is zero.
To understand why this is the case, let's break down the forces acting on the sled:
1. Gravity: The force of gravity acts vertically downwards and is given by the equation F_gravity = m * g, where m is the total mass of the sled and girl and g is the acceleration due to gravity (9.8 m/s^2).
- The component of gravity parallel to the incline (F_parallel) is: F_parallel = m * g * sin(theta), where theta is the angle of the incline (11° in this case).
- The component of gravity perpendicular to the incline (F_perpendicular) is: F_perpendicular = m * g * cos(theta).
2. Normal force: The normal force acts perpendicular to the incline and counteracts the component of gravity perpendicular to the incline. In this case, the normal force is equal to F_perpendicular.
3. Kinetic friction: The kinetic friction force opposes the motion of the sled and is given by the equation F_friction = μ * F_normal, where μ is the coefficient of kinetic friction between the sled and the snow runners.
- In this case, F_friction = μ * F_perpendicular.
4. Applied force: The father applies a horizontal force (F_applied) on the sled to move it up the hill. This force acts parallel to the incline.
Since the sled is moving up the hill with a constant velocity, we have the following equilibrium condition: F_applied = F_parallel + F_friction.
Now, let's consider the work done by each force:
- The work done by the applied force is given by the equation W_applied = F_applied * d, where d is the displacement of the sled (h = 3.9 m in this case). Since the sled is moving at a constant velocity, the net work done by the applied force is zero.
- The work done by the gravitational force is given by the equation W_gravity = F_parallel * d. However, since the sled is moving up the hill with a constant velocity, the net work done by the gravitational force is also zero.
- The work done by the friction force is given by the equation W_friction = F_friction * d. Again, since the sled is moving up the hill with a constant velocity, the net work done by the friction force is zero.
Therefore, the total work done on the sled is zero, and thus the answer to the question is zero.
Note: The presence of friction does not change the work done in this case because the work due to the applied force and the work due to gravity cancel out the work done by friction. If the sled were to move up the incline at an accelerating or decelerating rate, then the work done by friction would not be zero.