Work and energy of an object on a ice slope

In summary, the relationship between work and energy on an object on an ice slope is that as the object moves down the slope, work is done on it, resulting in an increase in kinetic energy. Friction plays a crucial role in this process, converting some of the object's kinetic energy into heat energy and causing it to slow down. The work done on an object on an ice slope can be negative when the object is moving up the slope and the force of gravity is acting against it. The mass of an object does not directly affect its work and energy on an ice slope, but a heavier object will have more energy on the slope compared to a lighter object. The work and energy of an object on an ice slope are not conserved
  • #1
ac7597
126
6
Homework Statement
A perfect hemisphere of frictionless ice has radius R=7 meters. Sitting on the top of the ice, motionless, is a box of mass m=7 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 11 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?

At some point, as the box slips down the ice and speeds up, it loses contact with the ice and flies off into the air. At what angle from the top does the box leave the ice? I'll provide the units for you.

(Hint: At the critical point, the component of the gravitational force pointing toward the center of the hemisphere is exactly equal to the force required to keep the box moving in a circular path around the center of the hemisphere)
Relevant Equations
KE=(1/2)mv^2
Homework Statement: A perfect hemisphere of frictionless ice has radius R=7 meters. Sitting on the top of the ice, motionless, is a box of mass m=7 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 11 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?

At some point, as the box slips down the ice and speeds up, it loses contact with the ice and flies off into the air. At what angle from the top does the box leave the ice? I'll provide the units for you.

(Hint: At the critical point, the component of the gravitational force pointing toward the center of the hemisphere is exactly equal to the force required to keep the box moving in a circular path around the center of the hemisphere)
Homework Equations: KE=(1/2)mv^2

The work of gravity on the box across= mg(distance) = mg(radius-radius(cos(theta)))
Thus work of gravity= (7kg)(9.8m/s^2)(7-7cos(11))= 8.8J

velocity = (2(8.8J)/(7kg))^(1/2)=1.59m/s

I am stuck on how to find the angle when the box slips
 

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  • #2
ac7597 said:
Homework Statement: A perfect hemisphere of frictionless ice has radius R=7 meters. Sitting on the top of the ice, motionless, is a box of mass m=7 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 11 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?

At some point, as the box slips down the ice and speeds up, it loses contact with the ice and flies off into the air. At what angle from the top does the box leave the ice? I'll provide the units for you.

(Hint: At the critical point, the component of the gravitational force pointing toward the center of the hemisphere is exactly equal to the force required to keep the box moving in a circular path around the center of the hemisphere)
Homework Equations: KE=(1/2)mv^2

I am stuck on how to find the angle when the box slips
The hint is in the hint.
 
  • #3
The component of the gravitational force pointing at the center is mg*cos(theta). The force needed to keep a circular path is m(a)=m(v^2 /radius). We don't know the velocity at the point the ball slips.
 
  • #4
What is the velocity as a function of theta?
 
  • #5
sin(theta)= velocity/radius ?
 
  • #6
ac7597 said:
sin(theta)= velocity/radius ?
That cannot be, it is dimensionally inconsistent. Consider how you solved the first part of the problem.
 
  • #7
If final velocity^2 = 2g(R-Rcos(theta))
velocity= 2g(R-Rcos(theta))^(1/2)
thus mgcos(theta)= (velocity)^2/R
finally theta= 48.2 degrees
 
  • #8
ac7597 said:
If final velocity^2 = 2g(R-Rcos(theta))
velocity= 2g(R-Rcos(theta))^(1/2)
thus mgcos(theta)= (velocity)^2/R
finally theta= 48.2 degrees
Some of those equations are still dimensionally inconsistent, but you have the right answer, so I assume they are just typos in posting your work.
You should have ##\frac{v^2}{rg}=2(1-\cos(\theta))## (energy) and ##\frac{v^2}{rg}=\cos(\theta)## (centripetal acceleration), whence ##\cos(\theta)=\frac 23##.
 

FAQ: Work and energy of an object on a ice slope

What is the relationship between work and energy on an ice slope?

The work done on an object on an ice slope is directly related to the energy of the object. As the object moves down the slope, it gains kinetic energy, which is a form of energy associated with motion. This increase in kinetic energy is a result of the work done on the object by the force of gravity.

How does the angle of the ice slope affect the work and energy of an object?

The angle of the ice slope affects the work and energy of an object by changing the amount of potential energy the object has. The steeper the slope, the greater the potential energy and the more work is required for the object to move up or down the slope. This also means that the object will have a higher kinetic energy as it moves down the slope.

What is the role of friction in work and energy on an ice slope?

Friction plays a significant role in the work and energy of an object on an ice slope. It acts as a resistance to the movement of the object and can reduce the amount of work done on the object. This results in a decrease in its kinetic energy and a slower rate of movement down the slope. The type and amount of friction also depend on the surface of the ice and the object's material.

How does the mass of an object affect its work and energy on an ice slope?

The mass of an object has a direct impact on its work and energy on an ice slope. A heavier object will require more work to move up or down the slope compared to a lighter object. This is because the work done by gravity is dependent on the mass of the object. Additionally, a heavier object will have more potential energy and therefore more kinetic energy as it moves down the slope.

What are some real-life applications of understanding work and energy on an ice slope?

Understanding the relationship between work and energy on an ice slope has many real-life applications. For example, it can be used in winter sports such as skiing and ice skating to improve performance and safety. It is also important in designing and building structures in icy environments, such as bridges or roads, to ensure they can withstand the forces of gravity and friction on an inclined surface.

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