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 71 kg, and the height of the water slide is 12.0 m. If the kinetic frictional force does -7.4 × 103 J of work, how fast is the student going at the bottom of the slide? Use g = 9.81 m/s2
Could I use the equation
vf=sqrt v0^2 + 2(g)(h0-hf) ?
No, not that equation.
final KE=initial PE - frictional loss
1/2 m vf^2=m*g*(12)-7.4E3
solve for vf
so for m
So for m I would plug in 71kg correct?
Yes, you can use the equation vf = sqrt(v0^2 + 2gh) to find the final velocity of the student at the bottom of the slide. Here's how you can apply the equation to solve the problem:
1. Identify the given information:
- Mass of the student (m): 71 kg
- Height of the water slide (h): 12.0 m
- Work done by the kinetic frictional force (W): -7.4 × 10^3 J
- Acceleration due to gravity (g): 9.81 m/s^2
2. Determine the initial velocity (v0):
Since the student starts from rest, the initial velocity is 0 m/s.
3. Find the final velocity (vf) using the equation:
vf = sqrt(v0^2 + 2gh)
Plug in the values:
vf = sqrt(0^2 + 2 * 9.81 * 12.0)
4. Solve for vf:
vf = sqrt(0 + 235.44)
vf ≈ sqrt(235.44)
vf ≈ 15.34 m/s
Therefore, the student will be going approximately 15.34 m/s at the bottom of the slide.