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 70 kg, and the height of the water slide is 12.2 m. If the kinetic frictional force does -7.9 × 103 J of work, how fast is the student going at the bottom of the slide?
To find the student's speed at the bottom of the slide, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system remains constant as long as no external forces are acting on it.
The initial mechanical energy of the student at the top of the slide is given by the potential energy (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 final mechanical energy of the student at the bottom of the slide is given by the kinetic energy (0.5mv^2), where v is the speed of the student.
Since the kinetic frictional force does negative work, it decreases the total mechanical energy of the system. Therefore, we can write the equation:
Initial potential energy - Work done by frictional force = Final kinetic energy
(mgh) - (-7.9 × 10^3 J) = 0.5mv^2
Now let's plug in the values:
mass (m) = 70 kg
acceleration due to gravity (g) = 9.8 m/s^2
height of the slide (h) = 12.2 m
work done by frictional force (-7.9 × 10^3 J)
(70 kg)(9.8 m/s^2)(12.2 m) + 7.9 × 10^3 J = 0.5(70 kg)v^2
Now we can solve for v, the speed of the student:
(8460 J) + 7.9 × 10^3 J = 0.5(70 kg)v^2
Take the sum of the left side:
16,260 J = 0.5(70 kg)v^2
Now, divide both sides by 0.5(70 kg):
(16,260 J) / (0.5(70 kg)) = v^2
Simplify:
584.14 J/kg = v^2
Finally, take the square root of both sides to find the speed:
v = √(584.14 J/kg)
v ≈ 24.16 m/s
Therefore, the student is going approximately 24.16 m/s at the bottom of the slide.