Solving Toboggan Homework: Work & Kinetic Energy

In summary, the conversation discusses the calculation of the vertical height a toboggan will reach on a frictionless icy hill, given its speed and the slope of the hill. The equation for calculating the vertical height is derived using the work and kinetic energy theorem. It is noted that the angle of the slope does not affect the vertical height in a frictionless scenario. The conversation also briefly discusses the work done by friction and gravity in the presence of friction, and clarifies that the change in potential energy is always the product of the weight and the vertical distance raised.
  • #1
Amar.alchemy
79
0

Homework Statement



At the base of a frictionless icy hill that rises at 25 degree above the horizontal, a toboggan has a speed of 12 m/s toward the hill. How high vertically above the base will it go before stopping?

Homework Equations



Work and Kinetic energy theorem

The Attempt at a Solution



W = −mgy, where y is the vertical height.
−mgy=−1/2 mv2 and
therefore y equals v2/2g

it means vertical height is independent of slope. Is this correct answer?? Bcoz if i assume that slope is zero then also it should reach the vertical height given by the above equation. Kindly clarify.
 
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  • #2
Amar.alchemy said:

Homework Statement



At the base of a frictionless icy hill that rises at 25 degree above the horizontal, a toboggan has a speed of 12 m/s toward the hill. How high vertically above the base will it go before stopping?

Homework Equations



Work and Kinetic energy theorem

The Attempt at a Solution



W = −mgy, where y is the vertical height.
−mgy=−1/2 mv2 and
therefore y equals v2/2g

it means vertical height is independent of slope. Is this correct answer?? Bcoz if i assume that slope is zero then also it should reach the vertical height given by the above equation. Kindly clarify.
That is indeed the correct answer. The angle of the slope would be important if the slope was frictionless, however, since the slope is indeed frictionless the vertical height should be independent of the inclination. The distance parallel to the slope will depend on the inclination.

If the slope was zero, your equation wouldn't make much sense now would it because it would never leave the horizontal.
 
  • #3
so, if i assume that there is some friction present , with kinetic friction co-efficient MUk, then wat is the work done by gravity??
 
  • #4
Amar.alchemy said:
so, if i assume that there is some friction present , with kinetic friction co-efficient MUk, then wat is the work done by gravity??
Well you'd first have to work out the work done by friction ...
 
  • #5
Work done by friction comes out to be -mukmgh / tan alpha

ok, so here comes the angle alpha, so what i am not understanding is why we are not considering alpha while calculating work done by gravity?? sorry if i am dragging this too much...
 
  • #6
Amar.alchemy said:
Work done by friction comes out to be -mukmgh / tan alpha
Correct.
Amar.alchemy said:
ok, so here comes the angle alpha, so what i am not understanding is why we are not considering alpha while calculating work done by gravity?? sorry if i am dragging this too much...
Okay, so if the hill has friction, then the kinetic energy can be transformed into the work done against gravity and work done against the friction. However, for the case of a frictionless hill, the kinetic energy is only transferred into gravitational potential energy. What is the only way to increase a body's gravitational potential energy?
 
  • #7
Then i have to raise its height against gravity... rite??

But still the work done by gravity in the "hill with friction is mgh"... rite??

what i mean is work done by gravity should be "mgh sin alpha"
 
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  • #8
Amar.alchemy said:
Then i have to raise its height against gravity... rite??
Correct.
Amar.alchemy said:
But still the work done by gravity in the "hill with friction is mgh"... rite??
Correct again, the change in potential energy of the body is always the product of the weight and the vertical distance raised.
Amar.alchemy said:
what i mean is work done by gravity should be "mgh sin alpha"
Why should that be the case?

Let's take an example. Can you answer the following questions:

(1) Suppose I have a mass m and I raise it vertically h meters. What is the change in gravitational potential energy of the mass?

(2) Suppose that once again I have a mass m and I raise it vertically h meters. Then I move it h meters horizontally. What is the change in gravitational potential energy of the mass?
 
  • #9
Ya, in both cases the answer is "mgh"

Thank you very much Hootenanny :-)
 
  • #10
Amar.alchemy said:
Ya, in both cases the answer is "mgh"

Thank you very much Hootenanny :-)
A pleasure :smile:
 

What is the definition of work and kinetic energy?

Work is defined as the transfer of energy that occurs when a force is applied to an object and the object is displaced in the direction of the force. Kinetic energy is the energy an object possesses due to its motion.

What is the formula for calculating work?

The formula for calculating work is W = Fd, where W is work, F is the force applied, and d is the distance the object is displaced in the direction of the force.

What is the relationship between work and kinetic energy?

Work and kinetic energy are directly related. The work done on an object is equal to the change in kinetic energy of the object. This means that if work is done on an object, its kinetic energy will increase, and if work is done by an object, its kinetic energy will decrease.

How is work and kinetic energy related to solving toboggan homework?

When solving toboggan homework, work and kinetic energy are used to calculate the energy needed to move the toboggan. The work done by the force of gravity on the toboggan is equal to its change in kinetic energy, and this information can be used to solve problems involving toboggan motion.

What are some real-world applications of work and kinetic energy?

Work and kinetic energy have many real-world applications, including calculating the amount of energy needed to move objects like cars, calculating the power output of machines, and understanding the mechanics of sports like skiing and snowboarding. They are also important in fields like engineering, physics, and mechanics.

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