Work of gravity on an object sliding down a frictionless sphere

In summary, we have a perfect hemisphere of frictionless ice with a radius of 6.5 meters and a box of mass 6 kg sitting motionless on top. As the box starts to slide down the sloping surface of the ice by an angle of 20 degrees, the work done by gravity is calculated to be 45.65J. Using the formula for kinetic energy, we can also determine that the box is moving at a speed of 2.77m/s. The question of what force is causing the box to slide down is then discussed, with the conclusion that the work done by gravity can be calculated using the formula -G*(mass of hemisphere)*(mass of box)/(x), where G is the gravitational constant and
  • #1
ac7597
126
6
Homework Statement
A perfect hemisphere of frictionless ice has radius R=6.5 meters. Sitting on the top of the ice, motionless, is a box of mass m=6 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 20 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?
Relevant Equations
g=9.8m/s^2
Homework Statement: A perfect hemisphere of frictionless ice has radius R=6.5 meters. Sitting on the top of the ice, motionless, is a box of mass m=6 kg.

The box starts to slide to the right, down the sloping surface of the ice. After it has moved by an angle 20 degrees from the top, how much work has gravity done on the box?

How fast is the box moving?
Homework Equations: g=9.8m/s^2

Force xy
normal force of sphere on box 0Fn
gravity mgsin(theta) -mgcos(theta)
totalm(ax) m(ay)=0

From the force diagram, Fn=mgcos(theta) and m(ax)=mgsin(theta). Net force=m(ax). Thus the integral of net force=mgsin(theta)*(distance traveled). Since the circumference of half a sphere = 2(pie)(6.5m)/2 = 20.42m, the distance traveled= (20.42m)*(20degree)/(180degree)= 2.27m. The final equation should be: work=(6kg)(9.8m/s^2)sin(20deg)*(2.27m)=45.65J.
 
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  • #2
Screen Shot 2019-10-03 at 1.31.47 AM.png
 
  • #3
You have a little more work to do to compute that integral. theta is not constant. It's certainly possible to do it this way, but there is an easier way to compute the work done by gravity that is valid for all paths, if you only know the starting point and the ending point.
 
  • #4
Like: integral of net force= -mgcos(theta) ?
 
  • #5
ac7597 said:
Like: integral of net force= -mgcos(theta) ?
Never integrate unless you have to.

Think energy!
 
  • #6
Is it kinetic energy = (1/2)mv^2 ?
 
  • #7
ac7597 said:
Is it kinetic energy = (1/2)mv^2 ?
That is the formula for KE. But the question is where did the KE come from?
 
  • #8
work done by gravity=∫ (G*(mass of hemisphere)*(mass of box) )/(x^2) dx. G=6.67*10^(-11) N*(m^2)/kg^2.
This becomes work= -(G*(mass of hemisphere)*(mass of box) )/ (x).
mass of ice hemisphere=(2/3)⋅π⋅(radius^3)⋅(mass density of ice) = (2/3)π (6.5m)^3 (0.92 g/cm^3) = 529159.39 kg.

Thus work1= -6.67*10^(-11) (529159.39kg)(6kg)/(6.5m) = -32.58 *10^(-6) J
work2= -6.67*10^(-11) (529159.39kg)(6kg)/(6.5m-2.27m) = -50.06 *10^(-6) J
work1-work2= -32.58 *10^(-6) J -(-50.06*10^(-6) J )= 17.48*10^(-6) J
 
  • #9
ac7597 said:
G*(mass of hemisphere)*(mass of box)
What force is causing the box to slide down? Indeed, what defines "down"?
 
  • #10
ac7597 said:
work done by gravity=∫ (G*(mass of hemisphere)*(mass of box) )/(x^2) dx. G=6.67*10^(-11) N*(m^2)/kg^2.
This becomes work= -(G*(mass of hemisphere)*(mass of box) )/ (x).
mass of ice hemisphere=(2/3)⋅π⋅(radius^3)⋅(mass density of ice) = (2/3)π (6.5m)^3 (0.92 g/cm^3) = 529159.39 kg.

Thus work1= -6.67*10^(-11) (529159.39kg)(6kg)/(6.5m) = -32.58 *10^(-6) J
work2= -6.67*10^(-11) (529159.39kg)(6kg)/(6.5m-2.27m) = -50.06 *10^(-6) J
work1-work2= -32.58 *10^(-6) J -(-50.06*10^(-6) J )= 17.48*10^(-6) J
Have you ever seen the term ##mgh##?
 
  • #11
is it work=mg(radius-radius(cos(theta)))? I got work=23.05J
 
  • #12
mass(velocity^2)(1/2)=23.05J. Thus velocity=2.77m/s
 
  • #13
ac7597 said:
mass(velocity^2)(1/2)=23.05J. Thus velocity=2.77m/s
Yes.
 

1. How does gravity affect an object sliding down a frictionless sphere?

Gravity is the force that pulls objects towards the center of the earth. As an object slides down a frictionless sphere, gravity acts on it by accelerating it towards the center of the sphere. The force of gravity increases as the object gets closer to the center of the sphere.

2. What is the work done by gravity on an object sliding down a frictionless sphere?

The work done by gravity is the product of the force of gravity and the distance the object travels. As the object slides down the frictionless sphere, gravity does positive work on it by accelerating it towards the center of the sphere. This work is equal to the change in the object's kinetic energy.

3. Does the weight of the object affect the work of gravity on a frictionless sphere?

No, the weight of the object does not affect the work of gravity on a frictionless sphere. The weight of an object is the force of gravity acting on it, and in this case, gravity is the only force acting on the object. Therefore, the weight of the object cancels out and does not affect the work done by gravity.

4. How does the radius of the sphere affect the work of gravity on an object sliding down?

The radius of the sphere does not directly affect the work of gravity on an object sliding down. However, the distance the object travels and the force of gravity both depend on the radius of the sphere. A larger radius will result in a longer distance traveled and a stronger force of gravity, which will result in more work being done on the object.

5. What is the relationship between the work of gravity and the speed of the object on a frictionless sphere?

The work done by gravity is directly proportional to the speed of the object on a frictionless sphere. As the object speeds up while sliding down the sphere, the work done by gravity also increases. This is because the force of gravity is acting on the object over a longer distance, resulting in more work being done on the object.

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