A 44.0 kgseal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.75 m higher than the surface of the water and the ramp is inclined at an angle of 35. degrees above the horizontal.
If the seal reaches the water with a speed of 4.00 {\rm m/s}, what is the work done by kinetic friction?
Work done AGAINST friction = (P.E loss) - (K.E. gain)
= M g H - (1/2)M V^2
(V is the speed when it hits the water)
The angle of the ramp doesn't matter when computing the answer, but it does affect how much frictional work is done
The work done BY the friction is negative
To find the work done by kinetic friction, we need to first find the work done by gravity and the work done by the net force on the seal.
1. Work done by gravity:
The work done by gravity can be calculated using the formula: W_gravity = m * g * h
where m is the mass of the seal (44.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical distance the seal slides down the ramp (given as 1.75 m).
W_gravity = 44.0 kg * 9.8 m/s^2 * 1.75 m
W_gravity = 752.6 J
2. Work done by the net force:
The work done by the net force can be calculated using the formula: W_net = m * a * d
where m is the mass of the seal (44.0 kg), a is the acceleration of the seal, and d is the distance the seal slides down the ramp (which we need to find).
To find the acceleration, we can use the equation: v^2 = u^2 + 2a*d
where v is the final velocity of the seal (4.00 m/s), and u is the initial velocity of the seal (which is 0 since the seal starts from rest).
Rearranging the equation, we have: a = (v^2 - u^2) / (2 * d)
Plugging in the values, we get: 4.00^2 = 0 + 2 * a * d
Simplifying, we get: 16 = 2a * d
We also know that the angle of the ramp is 35 degrees, so we can relate the distance d to the vertical height h as: d = h / sin(angle).
Plugging in the values, we get: 16 = 2a * (1.75 / sin(35))
Simplifying, we get: 16 = 2a * 3.23
Solving for a, we get: a = 16 / (2 * 3.23)
a = 2.48 m/s^2 (approximately)
Now we can find the distance d: d = h / sin(angle)
d = 1.75 m / sin(35)
d = 3.23 m (approximately)
Plugging in the values, we get: W_net = 44.0 kg * 2.48 m/s^2 * 3.23 m
W_net = 359.6 J (approximately)
3. Work done by kinetic friction:
The work done by kinetic friction can be calculated using the equation: W_friction = -μ_k * N * d
where μ_k is the coefficient of kinetic friction, N is the normal force, and d is the distance the seal slides down the ramp (3.23 m).
Since the seal is on an inclined plane, the normal force is given by: N = m * g * cos(angle)
Plugging in the values, we get: N = 44.0 kg * 9.8 m/s^2 * cos(35)
N = 359.1 N (approximately)
Now we can calculate the work done by kinetic friction: W_friction = -μ_k * N * d
However, the coefficient of kinetic friction (μ_k) is not provided in the question. It is necessary to have this information in order to calculate the work done by kinetic friction.
Therefore, we cannot determine the work done by kinetic friction without the coefficient of kinetic friction.