A skier of mass 70.4 kg, starting from rest, slides down a slope at an angle $\theta$ of 36.8° with the horizontal. The coefficient of kinetic friction is 0.14. What is the net work, i.e. net gain in kinetic energy, (in J) done on the skier in the first 8.4 s of descent?
See previous post.
To find the net work done on the skier, we need to calculate the change in kinetic energy.
First, let's find the gravitational force acting on the skier. The gravitational force can be calculated using the formula:
F_gravity = m * g * sin(θ)
where m is the mass of the skier, g is the acceleration due to gravity, and θ is the angle of the slope.
Given:
m = 70.4 kg
θ = 36.8°
We know that g is approximately 9.8 m/s^2.
Plugging in the values, we get:
F_gravity = (70.4 kg) * (9.8 m/s^2) * sin(36.8°)
Next, we can calculate the kinetic friction force using the formula:
F_friction = μ * N
where μ is the coefficient of kinetic friction and N is the normal force.
The normal force can be calculated using:
N = m * g * cos(θ)
Plugging in the values, we get:
N = (70.4 kg) * (9.8 m/s^2) * cos(36.8°)
Then, we can calculate the kinetic friction force:
F_friction = (0.14) * (N)
Now, we can calculate the net force acting on the skier:
F_net = F_gravity - F_friction
Once we have the net force, we can calculate the work done using the formula:
Work = F_net * d
where d is the displacement in the direction of the force.
Since the skier starts from rest, the initial velocity is zero. Using the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
We can rearrange the equation to solve for the final velocity:
v = u + at
v = 0 + (9.8 m/s^2) * (8.4 s) * sin(36.8°)
Now, we can find the displacement using the formula:
d = (1/2) * a * t^2
d = (1/2) * (9.8 m/s^2) * (8.4 s)^2
Finally, we can calculate the net work done on the skier using:
Work = F_net * d
Remember to convert the angles to radians when using trigonometric functions.
Plug in the known values and calculate the net work done on the skier in the first 8.4 s of descent.