At a playground, a 19.8 kg child plays on a slide that drops through a height of 2.13 m. The child starts at rest at the top of the slide. On the way down, the slide does a nonconservative work of -319 J on the child. What is the child's speed at the bottom of the slide?
Use conservation of energy and subtract 319 from the initial potential energy of MgH = 413 J. The difference is the kinetic energy at the bottom of the slide. Get the speed from that.
Well, well, well, looks like we have a kid on a slide situation here! Let's slide into solving this problem, shall we?
We can start by using the work-energy principle. The work done on the child by the slide equals the change in the child's kinetic energy. So, we have:
Work = ΔKE
Since the slide does a nonconservative work of -319 J on the child, we have:
-319 J = ΔKE
Now, the change in kinetic energy can be expressed as:
ΔKE = (1/2)mv^2
Where m is the mass of the child (19.8 kg) and v is their final velocity (what we're trying to find).
Substituting the values, we get:
-319 J = (1/2)(19.8 kg)v^2
Now it's time to get rid of that pesky fraction. Let's multiply both sides of the equation by 2:
-638 J = 19.8 kg * v^2
Dividing both sides by 19.8 kg, we have:
-638 J / 19.8 kg = v^2
Time to take the square root of both sides to find our velocity:
√(-638 J / 19.8 kg) = v
I'm sorry to say, but it seems like our velocity at the bottom of the slide involves an imaginary number. Physics can be a bit of a clown sometimes, but keep your spirit high and enjoy the slide nonetheless!
To find the speed of the child at the bottom of the slide, we can use the principle of conservation of mechanical energy. According to this principle, the initial mechanical energy of the child (consisting of gravitational potential energy) is equal to the final mechanical energy of the child (consisting of kinetic energy) plus the work done on the child.
The initial mechanical energy of the child is given by:
Initial mechanical energy = mgh
where m is the mass of the child (19.8 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the slide (2.13 m).
So, the initial mechanical energy of the child is:
Initial mechanical energy = 19.8 kg * 9.8 m/s^2 * 2.13 m
The final mechanical energy of the child is kinetic energy and can be calculated using the formula:
Final mechanical energy = (1/2)mv^2
where m is the mass of the child and v is the speed of the child at the bottom of the slide.
According to the principle of conservation of mechanical energy:
Initial mechanical energy = Final mechanical energy + Work done
So, we can write the equation as:
mgh = (1/2)mv^2 + Work done
Substituting the given values, we have:
19.8 kg * 9.8 m/s^2 * 2.13 m = (1/2)(19.8 kg)v^2 -319 J
Simplifying the equation, we get:
391.356 J = 9.9 kg*v^2 - 319 J
Rearranging the equation, we have:
9.9 kg*v^2 = 391.356 J + 319 J
9.9 kg*v^2 = 710.356 J
Dividing both sides of the equation by 9.9 kg and taking the square root of both sides, we can find the speed of the child at the bottom of the slide.
v = √(710.356 J / 9.9 kg)
Calculating this value, we find:
v ≈ 9.33 m/s
Therefore, the child's speed at the bottom of the slide is approximately 9.33 m/s.
To find the child's speed at the bottom of the slide, we can use the principle of conservation of mechanical energy.
The total mechanical energy of the child at the top of the slide is the sum of its potential energy (PE) and kinetic energy (KE). At the bottom of the slide, the total mechanical energy is the sum of the child's potential energy (which is now zero) and kinetic energy.
The potential energy of an object with mass (m) and height (h) is given by the equation PE = mgh, where g is the acceleration due to gravity (approximately 9.8 m/s²).
At the top of the slide, the potential energy (PE1) is given by:
PE1 = mgh
At the bottom of the slide, the total mechanical energy (E2) is given by:
E2 = KE2
Since the slide does nonconservative work on the child (-319 J), we need to subtract this work from the total mechanical energy at the bottom of the slide to find the child's final kinetic energy.
KE2 = E2 - (-319 J)
KE2 = E2 + 319 J
KE2 = PE2 + 319 J (since PE2 is zero at the bottom of the slide)
Now, we can equate the initial potential energy to the final kinetic energy:
PE1 = KE2 + 319 J
Using the equation for potential energy, we have:
mgh = KE2 + 319 J
Solving for KE2 (the kinetic energy at the bottom of the slide), we get:
KE2 = mgh - 319 J
Finally, we can find the child's speed (v2) at the bottom of the slide using the equation for kinetic energy:
KE2 = (1/2)mv2²
Rearranging the equation, we have:
v2² = (2KE2) / m
Substituting the expression for KE2, we get:
v2² = (2(mgh - 319 J)) / m
Simplifying, we find:
v2² = 2gh - (638 J/m)
Taking the square root of both sides, we get:
v2 = √(2gh - (638 J/m))
Now we can plug in the values given in the question:
m = 19.8 kg (mass of the child)
g = 9.8 m/s² (acceleration due to gravity)
h = 2.13 m (height of the slide)
Substituting these values into the equation for v2, we can calculate the child's speed at the bottom of the slide.