Stuck on some Physics, help please (Work and Energy)?

In summary, the conversation discusses two physics problems - one involving a skeleton racer and the other involving a ski lift. The first problem asks for the speed of a rider at the bottom of a track with a total height of 104m, assuming no nonconservative forces. The second problem asks for the average power provided by the tension in a cable pulling four skiers at a constant speed to a height of 140m in 2.0min. The person seeking help understands the concepts but is struggling with calculating the work involved in the second problem.
  • #1
Marcelgluch
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0
Stuck on some Physics, help please! (Work and Energy)?

K this first one I am not sure how to do it.. i did the second part, that's because i had all the variables but in this one i dont...

ok this is about skeleton racers... the track total height is 104m, in the absence of nonconservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? ignore initial velocity since its relatively small.

and

In 2.0min a ski lift raises four skiers at constant speed to a height of 140m. the average mass of each skier is 65kg. what is the average power provided by the tension in the cable pulling...

i know what to but i just can't figure out the Work... i know that power = work/time.. so if i could get some help that would be greatly appreciated!
 
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  • #2


can someone please help me understand these two problems??
 
  • #3


Sure, I'd be happy to help with these physics questions! For the first question about the skeleton racer, we can use the conservation of energy principle to solve for the speed at the bottom of the track. Since there are no nonconservative forces acting on the rider, we can equate the initial potential energy (mgh) at the top of the track to the final kinetic energy (1/2mv^2) at the bottom of the track. So, we have:

mgh = 1/2mv^2

We can cancel out the mass and rearrange to solve for v:

v = √(2gh)

Substituting in the values given, we get:

v = √(2*9.8m/s^2*104m) = 45.8m/s

For the second question about the ski lift, we can use the equation for power, P = W/t, where W is the work done and t is the time. We can calculate the work done by the ski lift by using the formula W = mgh, where m is the mass of the skiers, g is the acceleration due to gravity, and h is the height they were lifted. So, we have:

W = (4*65kg)*9.8m/s^2*140m = 352,800J

Now, we can plug this value into the power equation:

P = (352,800J)/(2min*60s/min) = 2,940W

So, the average power provided by the tension in the cable is 2,940 watts. I hope this helps!
 

Related to Stuck on some Physics, help please (Work and Energy)?

1. What is work in physics?

In physics, work is defined as the application of a force over a distance. It is measured in joules (J) and is calculated by multiplying the force applied by the distance over which it is applied.

2. How is work related to energy?

Work and energy are closely related concepts in physics. Work is the transfer of energy from one object to another, or the conversion of energy from one form to another. Work is also a way of changing an object's energy level.

3. Can you explain the difference between kinetic and potential energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object has due to its position or state. Kinetic energy is dependent on an object's mass and velocity, while potential energy is dependent on an object's position and the forces acting upon it.

4. What is the work-energy theorem?

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In other words, the net work done on an object is equal to the change in its velocity.

5. Can you provide an example of calculating work and energy in a real-life scenario?

One example is lifting a book off the ground and placing it on a shelf. In this scenario, the work done is equal to the force applied (lifting the book) multiplied by the distance it is lifted. This work is then converted into potential energy as the book is raised to a higher position on the shelf.

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