at the moment when a shot putter releases a 4kg shot the shot is 1.25 m above the ground and travelling at 7 m/s. it reaches a maxium height of 3.7 m above the ground and then falls to the ground. air resistance is negligible and the shot goes straight up and then comes back down
What was the potential energy of the shot as it left the hand relative to the ground?
What was the kinetic energy of the shot as it left the hand?
What was the total energy of the shot as it reached its maximum height?
What was the kinetic energy of the shot just as it struck the ground?
typo
147.05 J for #4
Take potential = 0 at ground level as instructed.
1. m g h = 4 * 9.81 * 1.25 =49.05 J
2. (1/2) m v^2 = .5*4*49 = 98 J
3. No friction so PE + Ke = constant
= 49.05 + 98 = 147.05 J
by the way that means when it stops at the top it should have potential energy of 147 J
147 = m g h = 4*9.81 * h
h = 3.74, well they are a little off, maybe took 9.8 or 10 for g instead of 9.81
4. Potential energy lower by 49.05 because the ground is lower so it speeds up. All the energy is kinetic now
17.05
To answer your questions, we need to use the principles of potential energy, kinetic energy, and conservation of energy.
1. Potential Energy of the shot as it left the hand relative to the ground:
The potential energy (PE) of an object is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
In this case, the mass of the shot (m) is 4 kg, and the height (h) is 1.25 m. The acceleration due to gravity (g) is approximately 9.8 m/s^2.
So, the potential energy is calculated as follows:
PE = mgh = 4 kg * 9.8 m/s^2 * 1.25 m = 49 J.
2. Kinetic Energy of the shot as it left the hand:
The kinetic energy (KE) of an object is given by the formula KE = (1/2)mv^2, where m is the mass and v is the velocity.
In this case, the mass of the shot (m) is 4 kg, and the velocity (v) is 7 m/s.
So, the kinetic energy is calculated as follows:
KE = (1/2)mv^2 = 0.5 * 4 kg * (7 m/s)^2 = 98 J.
3. Total Energy of the shot as it reached its maximum height:
At the maximum height, the potential energy reaches its maximum value, while the kinetic energy is zero (as the shot comes to a stop momentarily). Therefore, the total energy (E) of the shot at its maximum height is equal to the potential energy.
So, the total energy is 49 J (as calculated in question 1).
4. Kinetic Energy of the shot just as it struck the ground:
When the shot reaches the ground, the potential energy is zero, and all its initial potential energy is converted into kinetic energy.
Therefore, the kinetic energy just as it strikes the ground is equal to the initial potential energy.
So, the kinetic energy is 49 J.
To find the potential energy of the shot as it left the hand relative to the ground, we can use the formula for gravitational potential energy:
Potential energy = mass * gravity * height
Given that the mass of the shot is 4kg, the height is 1.25m, and gravity is approximately 9.8 m/s^2, we can calculate the potential energy:
Potential energy = 4 kg * 9.8 m/s^2 * 1.25 m = 49 J
So, the potential energy of the shot as it left the hand relative to the ground is 49 Joules.
To find the kinetic energy of the shot as it left the hand, we can use the formula for kinetic energy:
Kinetic energy = 0.5 * mass * velocity^2
Given that the mass of the shot is 4kg and the velocity is 7 m/s, we can calculate the kinetic energy:
Kinetic energy = 0.5 * 4 kg * (7 m/s)^2 = 98 J
So, the kinetic energy of the shot as it left the hand is 98 Joules.
The total energy of the shot as it reached its maximum height remains the same, as energy is conserved in an isolated system (ignoring air resistance). Therefore, the total energy of the shot as it reached its maximum height is the sum of the potential and kinetic energy:
Total energy = potential energy + kinetic energy = 49 J + 98 J = 147 J
So, the total energy of the shot as it reached its maximum height is 147 Joules.
When the shot strikes the ground, it will have all of its initial potential energy converted into kinetic energy. Therefore, the kinetic energy of the shot just as it struck the ground would be equal to the potential energy that it had when it left the hand:
Kinetic energy = 49 J
So, the kinetic energy of the shot just as it struck the ground is 49 Joules.