1) 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 × 10^3 J of work, how fast is the student going at the bottom of the slide? Use g = 9.81 m/s2
Would I do sqrt (2)(-7.4e^3)+1/2(12)(71) - (71)(9.8)?
Well, let me break it down for you. You're on the right track with the equation, but it needs a little bit of clown magic to make it right. Here's the corrected equation:
v = sqrt(2gh - 2*(-7.4e^3) / m)
First, let's calculate the potential energy at the top of the slide:
PE = mgh = 71 * 9.81 * 12 = 8409.12 J
Now, taking into account the work done by friction:
Work = -7.4e^3 J
So, we subtract the work done by friction from the initial potential energy:
New PE = 8409.12 J - (-7.4e^3) J
Next, we use the conservation of mechanical energy to find the final kinetic energy:
New PE = KE
Finally, we use the equation for kinetic energy to solve for the velocity:
KE = 1/2 * m * v^2
v = sqrt(2 * New PE / m)
Now, plug in the values and let the magic happen!
To find the speed of the student at the bottom of the slide, you can use the conservation of mechanical energy. The total mechanical energy at the top of the slide (which includes potential energy and kinetic energy) should be equal to the total mechanical energy at the bottom of the slide (which includes only kinetic energy).
1. Calculate the potential energy at the top of the slide:
Potential Energy = mass * gravity * height
Potential Energy = 71 kg * 9.81 m/s^2 * 12.0 m
2. Calculate the work done by the frictional force:
Work = -7.4 × 10^3 J
3. Use the conservation of mechanical energy to set up the equation:
Potential Energy at the top = Kinetic Energy at the bottom - Work by friction
71 kg * 9.81 m/s^2 * 12.0 m = (1/2) * mass * velocity^2 - 7.4 × 10^3 J
4. Solve the equation for the velocity:
velocity^2 = (2 * (71 kg * 9.81 m/s^2 * 12.0 m + 7.4 × 10^3 J)) / 71 kg
velocity = √[((2 * (71 kg * 9.81 m/s^2 * 12.0 m + 7.4 × 10^3 J)) / 71 kg]
So, you would calculate the square root of the expression you provided: sqrt(2*(-7.4e^3) + (1/2)*(12)*(71) - (71)*(9.8)).
To find the speed of the student at the bottom of the slide, you can use the principle of conservation of energy. The total initial energy at the top of the slide will equal the total final energy at the bottom of the slide.
The initial energy at the top of the slide is given by the potential energy, which is the product of mass (m), gravitational acceleration (g), and height (h): mgh.
In this case, the mass (m) is 71 kg, the gravitational acceleration (g) is 9.81 m/s^2, and the height (h) is 12.0 m. So, the initial energy is 71 × 9.81 × 12.0 = 8381.32 J.
The final energy at the bottom of the slide is the sum of the kinetic energy (K) and the work done by the kinetic frictional force (W_friction). The kinetic energy is given by 0.5mv^2, where v is the speed at the bottom. In this case, m is 71 kg.
So, the equation becomes:
8381.32 J = 0.5 × 71 × v^2 - 7.4 × 10^3 J
To solve for v, rearrange the equation:
0.5 × 71 × v^2 = 8381.32 J + 7.4 × 10^3 J
0.5 × 71 × v^2 = 15781.32 J
Now, divide by 0.5 × 71 to isolate v^2:
v^2 = (15781.32 J) / (0.5 × 71)
v^2 ≈ 444.6
Now, take the square root of both sides to solve for v:
v ≈ √(444.6)
v ≈ 21.1 m/s
Therefore, the student is going approximately 21.1 m/s at the bottom of the slide.
m•g•h- W(fr) = m•v²/2
v = sqrt{2•g•h – [2•W(fr)/m]} =
sqrt{2•9.81•12 – [2•7.4•10³/71]} = 11.44 m/s