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 74 kg, and the height of the water slide is 12.6 m. If the kinetic frictional force does -6.1 × 103 J of work, how fast is the student going at the bottom of the slide?
To find the speed of the student at the bottom of the slide, we can use the principle of conservation of mechanical energy. The initial mechanical energy is given by the potential energy at the top of the slide, and the final mechanical energy is given by the kinetic energy at the bottom of the slide.
The potential energy at the top of the slide is given by the formula:
PE = mgh
where m is the mass of the student, g is the acceleration due to gravity, and h is the height of the slide.
The kinetic energy at the bottom of the slide is given by the formula:
KE = 0.5mv^2
where m is the mass of the student, and v is the velocity of the student at the bottom of the slide.
Since there is a frictional force acting on the student, the work done by friction needs to be taken into account. The work done by friction is given by the formula:
Work = -F_friction * d
where F_friction is the frictional force, and d is the distance traveled.
To find the velocity at the bottom of the slide, we need to set the initial mechanical energy equal to the final mechanical energy minus the work done by friction:
PE = KE - Work
Substituting the formulas for potential energy and kinetic energy, we get:
mgh = 0.5mv^2 - Work
Rearranging the equation and solving for v:
v^2 = (2 * (mgh + Work)) / m
v = √((2 * (mgh + Work)) / m)
Plugging in the given values:
m = 74 kg
g = 9.8 m/s^2
h = 12.6 m
Work = -6.1 × 10^3 J
v = √((2 * (74 kg * 9.8 m/s^2 * 12.6 m - (-6.1 × 10^3 J))) / 74 kg)
Simplifying the equation:
v = √((2 * (74 kg * 9.8 m/s^2 * 12.6 m + 6.1 × 10^3 J)) / 74 kg)
Calculating the value:
v ≈ √((2 * (873.36 kg·m^2/s^2 + 6.1 × 10^3 J)) / 74 kg)
v ≈ √((2 * (873.36 kg·m^2/s^2 + 6.1 × 10^3 kg·m^2/s^2)) / 74 kg)
v ≈ √((2 * 7.9736 × 10^3 kg·m^2/s^2) / 74 kg)
v ≈ √(215.02 kg·m^2/s^2)
v ≈ 14.65 m/s
Therefore, the student is moving at approximately 14.65 m/s at the bottom of the slide.