A father pushes horizontally on his daughter's sled to move it up a snowy incline, as illustrated in the figure, with h = 3.7 m and θ = 18°. 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?
Well, that's quite a slippery situation! Let's calculate the work done by the father. Since the sled moves up the hill with a constant velocity, we know that the sum of the forces acting on it must be zero.
So, the force the father applies must counteract two forces: the force of gravity and the force of friction. We can break down the forces into components parallel and perpendicular to the hill.
The perpendicular component of the force of gravity is mg * cos(theta), where m is the mass (35 kg) and theta is the angle of the incline (18°). The parallel component of the force of gravity is mg * sin(theta).
Since the sled is moving at a constant velocity, the father must apply a force equal in magnitude and opposite in direction to the sum of these two components. The parallel component of the force of gravity is counteracted by the frictional force, so we can write:
Force applied by father = mg * cos(theta) + frictional force
Now let's find the frictional force. The frictional force is given by the coefficient of kinetic friction times the normal force. The normal force is equal to the perpendicular component of the force of gravity, which is mg * cos(theta). So we have:
Frictional force = μ * normal force
= μ * mg * cos(theta)
Substituting this into our previous equation:
Force applied by father = mg * cos(theta) + μ * mg * cos(theta)
= mg * (1 + μ * cos(theta))
Now we can calculate the work done by the father. Since work is force multiplied by distance, and the sled is moved horizontally, the work is simply:
Work = force applied by father * distance
= mg * (1 + μ * cos(theta)) * h
Plugging in the given values:
Work = 35 kg * (1 + 0.20 * cos(18°)) * 3.7 m
And I'm afraid the punchline is... drum roll, please... the work done by the father is approximately 522 Joules. Phew, that's some heavy lifting!
To calculate the work done by the father in moving the sled from the bottom to the top of the hill, we need to determine the net force acting on the sled and multiply it by the distance it is moved.
First, let's find the force of gravity acting on the sled. The force of gravity can be calculated using the equation:
F_gravity = m * g
Where:
m = mass of sled and girl = 35 kg
g = acceleration due to gravity = 9.8 m/s^2
F_gravity = 35 kg * 9.8 m/s^2
F_gravity = 343 N
Next, let's find the force of friction acting on the sled. The force of friction can be calculated using the equation:
F_friction = coefficient of friction * normal force
The normal force is the force perpendicular to the surface of contact, which is equal to the component of gravity parallel to the incline. Since the angle of the incline is given as θ = 18°, the normal force can be calculated as:
F_normal = m * g * cos(θ)
F_normal = 35 kg * 9.8 m/s^2 * cos(18°)
F_normal = 329.38 N
The force of friction can now be calculated as:
F_friction = 0.20 * F_normal
F_friction = 0.20 * 329.38 N
F_friction = 65.876 N
Since the sled is moving up the hill at a constant velocity, the net force acting on it is zero. Therefore, the force exerted by the father must cancel out the force of gravity and the force of friction. The net force can be calculated as:
F_net = F_gravity + F_friction
F_net = 343 N + 65.876 N
F_net = 408.876 N
Finally, we can calculate the work done by the father by multiplying the net force by the distance the sled is moved, h = 3.7 m:
Work = F_net * h
Work = 408.876 N * 3.7 m
Work ≈ 1512.157 Joules
Therefore, the father does approximately 1512.157 Joules of work in moving the sled from the bottom to the top of the hill.
To determine the work done by the father in moving the sled from the bottom to the top of the hill, we need to calculate the total work done against the force of gravity and the force of friction.
1. Work done against the force of gravity:
The force of gravity acting on the sled and the girl can be calculated using the equation:
F_gravity = m * g
where m is the total mass (35 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the sled moves up the hill with a constant velocity, the gravitational force does not contribute to the work done. This is because the displacement of the sled is perpendicular to the gravitational force. Therefore, the work done against gravity is zero.
2. Work done against the force of friction:
The force of friction opposing the motion of the sled can be calculated using the equation:
F_friction = μ * F_normal
where μ is the coefficient of kinetic friction (0.20) and F_normal is the normal force.
The normal force can be calculated as:
F_normal = m * g * cos(θ)
where θ is the angle of incline (18°).
The work done against friction can be calculated using the equation:
Work_friction = F_friction * d
where d is the displacement of the sled up the hill (h = 3.7 m).
Now let's calculate the required values and substitute them into the equations:
F_normal = 35 kg * 9.8 m/s^2 * cos(18°)
F_normal ≈ 323.25 N
F_friction = 0.20 * 323.25 N
F_friction ≈ 64.65 N
Work_friction = 64.65 N * 3.7 m
Work_friction ≈ 239.36 J
Therefore, the work done by the father in moving the sled from the bottom to the top of the hill is approximately 239.36 Joules (J).