A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 75 kg, and the height of the water slide is 11.1 m. If the kinetic frictional force does -6.8 × 103 J of work, how fast is the student going at the bottom of the slide?
To find the velocity of the student at the bottom of the slide, we can make use of the principle of conservation of energy, which states that the total mechanical energy of a system is conserved if only conservative forces are acting.
In this scenario, the initial mechanical energy of the student at the top of the slide consists of potential energy due to their height above the ground. At the bottom of the slide, all the potential energy gets converted into kinetic energy.
First, let's find the initial potential energy (U) of the student at the top of the slide using the formula:
U = mgh
Where,
m = mass of the student = 75 kg
g = acceleration due to gravity = 9.8 m/s²
h = height of the slide = 11.1 m
Substituting the values into the equation, we get:
U = (75 kg)(9.8 m/s²)(11.1 m)
U ≈ 8,093.35 J
Next, we need to determine the final kinetic energy (K) of the student at the bottom of the slide. Since all the potential energy at the top is converted into kinetic energy at the bottom, we can equate the two:
K = U - Wfr
Where,
K = final kinetic energy
Wfr = work done by frictional force = -6.8 × 10^3 J (since it does negative work)
Substituting the values, we get:
K = 8,093.35 J - (-6.8 × 10^3 J)
K = 8,093.35 J + 6.8 × 10^3 J
K = 14,893.35 J
Finally, the kinetic energy is related to the mass (m) and velocity (v) of an object by the formula:
K = (1/2)mv²
Rearranging the equation to solve for v, we have:
v = √(2K/m)
Substituting the values, we get:
v = √(2(14,893.35 J)/(75 kg))
v ≈ √397.24 m²/s²
v ≈ 19.93 m/s (rounded to two decimal places)
Therefore, the student is going approximately 19.93 m/s at the bottom of the slide.
To find the speed of the student at the bottom of the slide, we need to use the principle of conservation of mechanical energy.
The total mechanical energy at the top of the slide is given by the sum of the gravitational potential energy and the initial kinetic energy:
E_top = m * g * h
where:
m = mass of the student = 75 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the water slide = 11.1 m
E_top = 75 kg * 9.8 m/s^2 * 11.1 m
= 819.645 J (rounded to three decimal places)
The work done by the kinetic frictional force is equal to the change in mechanical energy, which is given as -6.8 × 10^3 J.
Work_done = -6.8 × 10^3 J
Since work done is defined as the force applied multiplied by the distance over which it is applied, we can use the equation:
Work_done = Force * Distance
In this case, the frictional force acts parallel to the direction of motion, so the work_done is negative (since it is done against the motion).
Now, let's assume the distance traveled by the student is the same as the height of the water slide, which is 11.1 m.
Therefore, we can calculate the force of kinetic friction:
Force = Work_done / Distance
= -6.8 × 10^3 J / 11.1 m
= -612.612 N (rounded to three decimal places)
The net work done on the student is the sum of the work done by gravity and the work done by the kinetic frictional force:
Net_work_done = Work_done_gravity + Work_done_friction
Since the work done by gravity is equal to the change in gravitational potential energy, we can write:
Net_work_done = mgΔh + (-6.8 × 10^3 J)
The final kinetic energy at the bottom of the slide is given by:
E_bottom = 1/2 * m * v^2
where v is the speed of the student at the bottom of the slide.
According to the principle of conservation of mechanical energy, the total mechanical energy at the top of the slide (E_top) must be equal to the sum of the final kinetic energy at the bottom of the slide (E_bottom) and the work done by the kinetic frictional force (Work_done_friction):
E_top = E_bottom + Work_done_friction
Substituting the equations and values:
819.645 J = (1/2 * 75 kg * v^2) + (-6.8 × 10^3 J)
Rearranging the equation to solve for v:
1/2 * 75 kg * v^2 = 819.645 J + 6.8 × 10^3 J
v^2 = [2 * (819.645 J + 6.8 × 10^3 J)] / 75 kg
v^2 = 191.259 J / kg
Taking the square root of both sides:
v ≈ √(191.259 J / kg)
v ≈ 13.842 m/s
Therefore, the student will be going at approximately 13.842 m/s at the bottom of the slide.