Energy related problem

  • Thread starter alevis
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In summary, we have a rider on a water slide with a total mass of 80.0 kg starting at the top of the slide with a speed of 3.0 m/s. The slide is 10 m high, 55 m long, and 0.50 m wide with negligible friction. The rider then skims across the water for 50 m before coming to rest. To find the magnitude of the force the water exerts on the sled, we use the equation PE = mgh and KE = 1/2mv2. By setting these equal to each other and considering the initial kinetic energy from the rider's velocity, we can find the total kinetic energy at the bottom of the slide. Using the
  • #1
alevis
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Homework Statement


Yhou built a water slide . A rider on a small sled, of total mass 80.0 kg, pushed off to start at the top of the slide (point A) with a speed of 3.0 m/s. The chute was 10 m high at the top, 55 m long, and 0.50 m wide. Along its length, 725 wheels made friction negligible. Upon leaving the chute horizontally at its bottom end (point B), the rider skimmed across the water of Long Island Sound for as much as 50 m, “skipping along like a flat pebble,” before at last coming to rest and swimming ashore, pulling his sled after him.
(c) Find the magnitude of the force the water exerts on the sled.
(d) Find the magnitude of the force the chute exerts on the sled at point C just before leaving the slide.

Homework Equations


PE = mgh
KE = 1/2mv2


The Attempt at a Solution


PE = KE
mgh = 1/2mv2
Stuck!
I don't know were to go from here.
 
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  • #2
You've done well so far, but are missing one component in terms of energy.

Remember, the rider's initial velocity is given as 3.0 m/s (in the direction of the slide). Your "initial energy" equation will need both the potential energy (mgh) plus the kinetic energy from the initial velocity.

When the rider reaches the bottom, all potential energy should become kinetic energy, as you noted.

Now, once you have the total kinetic energy, what can you find using your equation KE = (1/2) m v2? (Hint - you know m)

At this point, you are looking for a force that will bring the rider to a stop in 50 meters. This means there must be some force acting on the rider to make him go from his initial velocity to his final velocity (zero).

Good luck.
 
  • #3


I would approach this problem by breaking it down into smaller parts and using the relevant equations to solve for each part.

First, I would calculate the potential energy (PE) of the rider at the top of the slide using the equation PE = mgh, where m is the mass of the rider (80.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the slide (10 m). This would give us a value for the potential energy of the rider at point A.

Next, I would use the equation for kinetic energy (KE) to calculate the initial kinetic energy of the rider at point A. This would give us a value for the initial kinetic energy of the rider.

Since the problem states that friction is negligible, we can assume that the total mechanical energy (PE + KE) of the rider remains constant throughout the slide. Therefore, at point B, the rider's kinetic energy is equal to the initial potential energy at point A.

To calculate the force exerted by the water on the sled, we can use the equation F = ma, where m is the mass of the sled (80.0 kg) and a is the acceleration of the sled. Since the sled is coming to a stop, the acceleration (a) is negative and equal to the deceleration of the sled.

To calculate the deceleration of the sled, we can use the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (3.0 m/s), and s is the distance traveled (50 m). Solving for a, we get a value of -0.36 m/s^2.

Substituting this value into the equation F = ma, we get a value for the force exerted by the water on the sled, which is approximately 28.8 N.

To calculate the force exerted by the chute on the sled at point C, we can use the equation F = ma, where m is the mass of the sled (80.0 kg) and a is the acceleration of the sled at point C. To find the acceleration at point C, we can use the equation v^2 = u^2 + 2as, where v is the final velocity (3.0 m/s), u is the initial velocity (0
 

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