a small rock with mass 0.20 kg is released from rest at point A which is at the top edge of a large, hemispherical bowl with radius R = .5m. assume that the size of the rock is small compared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls.The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has a magnitude 0.22 J.
A. Between points A and B, how much work is done on the rock by (i) the normal force and (ii) gravity?
B. What is the speed of the rock as it reaches point B?
C. Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain.
D. Just as the rock reaches point B, what is the due to the normal force on it due to the bottom of the bowl?
Between points A and B, how much work is done on the rock by the normal force?
-There is no work done by the normal force because all of the work is being done by gravity.
W=0J
Between points A and B, how much work is done on the rock by gravity?
-Work done by gravity is just the gravitational potential energy lost by moving in the vertical direction, so you use the equation Wg=mgh. In this case, h is just the radius so I'll sub that instead
Wg=mgR
Wg=(.26)*(9.8)*(.50)
Wg=1.274 J
What is the speed of the rock as it reaches point B?
-Find the speed by using the equation W=(.5)mv^2. For this part, you have to take into account the friction, Wt=Wg-Wf. So, solving for v:
v=sqrt((Wg-Wf)/(.5*m))
v=sqrt((1.274-.22)/(.5*.26))
v=2.85
Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl?
-When any object reaches the bottom of a loop or circle, the normal force is equal to the weight(mg) added to the mass times the radial acceleration(Arad). The equation looks like this: n=mg+m(Arad). First, you need to find the radial acceleration which is equal to v^2/R.
Arad=v^2/R
Arad=(2.85^2)/.5
Arad= 16.25
Now, you just plug in everything you know:
n=mg+m(Arad)
n=(.26*9.8)+(.26*16.25)
n=6.77N
hope that help ;)
Why are you using 0.26 when the mass in the problem says 0.20 kg?
Is it possible to have B further explained?
A.
(i) The work done by the normal force on the rock between points A and B is zero. This is because the normal force is perpendicular to the displacement of the rock, so the work done is zero. It's like getting paid for a job interview - it just doesn't happen.
(ii) The work done by gravity on the rock between points A and B can be calculated using the formula W = mgh, where m is the mass (0.20 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the change in height. In this case, the change in height is equal to the radius of the bowl (0.5 m). So, the work done by gravity is W = (0.20 kg)(9.8 m/s^2)(0.5 m) = 0.98 J.
B. To determine the speed of the rock as it reaches point B, we can use the principle of conservation of energy. The total energy at point A, when the rock is at rest, is solely gravitational potential energy, given by PE = mgh. When the rock reaches point B, all of the potential energy is converted into kinetic energy, given by KE = (1/2)mv^2, where v is the speed of the rock. Equating these two energies, we have mgh = (1/2)mv^2. Solving for v, we get v = sqrt(2gh), where g is the acceleration due to gravity and h is the change in height. Plugging in the values, v = sqrt(2 * 9.8 m/s^2 * 0.5 m) = 3.94 m/s (approximately).
C. Of the three forces acting on the rock as it slides down the bowl (friction, normal force, and gravity), only gravity is constant. The force of gravity is always pointing straight down towards the center of the Earth, so its magnitude doesn't change. However, both the normal force and the frictional force depend on the position and motion of the rock. The normal force changes as the rock moves up or down the slope of the bowl, while the frictional force opposes the motion of the rock, so its magnitude may vary depending on the speed of the rock.
D. Just as the rock reaches point B, the normal force due to the bottom of the bowl is equal in magnitude but opposite in direction to the weight of the rock. This is because at point B, the rock is no longer accelerating and its velocity is momentarily zero. So, the normal force precisely balances the gravitational force to prevent the rock from sinking into the bottom of the bowl. In other words, the normal force at point B is equal to the weight of the rock.
To solve this problem, we will use the concepts of work, energy, and forces.
A. Let's calculate the work done on the rock by (i) the normal force and (ii) gravity between points A and B:
(i) The work done by the normal force can be calculated using the equation W = F * d * cos(theta), where W is the work done, F is the force, d is the displacement, and theta is the angle between the force and displacement vectors.
In this case, the normal force is perpendicular to the displacement, so the angle between them is 90 degrees. Since cos(90) = 0, the work done by the normal force is 0 J.
(ii) The work done by gravity can be calculated using the equation W = m * g * h, where W is the work done, m is the mass, g is the acceleration due to gravity, and h is the vertical height.
In this case, the distance between points A and B is equal to the height of the hemisphere, which is R. Therefore, the work done by gravity is given by W = m * g * R.
B. To find the speed of the rock at point B, we can use the principle of conservation of energy, which states that the total mechanical energy of a system remains constant if no external forces are acting on it.
The mechanical energy at point A is given by the sum of its potential energy and kinetic energy: E_A = m * g * R + 0 (since it's at rest).
At point B, the mechanical energy is given by the sum of the potential energy (which is 0 at the bottom of the bowl) and the kinetic energy: E_B = 0 + (1/2) * m * v^2, where v is the speed of the rock at point B.
Since mechanical energy is conserved, we can equate E_A and E_B: m * g * R = (1/2) * m * v^2.
By canceling out the mass and solving for v, we get v = √(2 * g * R).
C. The three forces acting on the rock as it slides down the bowl are friction, gravity, and the normal force.
The force of gravity is constant throughout the motion as it acts vertically downwards.
The normal force is perpendicular to the surface and changes as the rock moves from point A to point B. It is largest at point A (equal to the weight of the rock) and decreases as the rock moves down.
The force of friction opposes the motion and is responsible for doing work on the rock. Its magnitude can be determined using the work done by friction given in the problem statement.
D. Just as the rock reaches point B, the normal force on it due to the bottom of the bowl is equal in magnitude but opposite in direction to the weight of the rock. This is because the rock is in equilibrium at point B, and the net force acting on it is zero.