At the moment when a shot-putter releases a 7.26-kg shot, the shot is 2.0 m above the ground and travelling at 15.0 m/s. It reaches a maximum height of 8.0 m above the ground and then falls to the ground. Assume that air resistance is negligible.
(a) What is the total mechanical energy of the shot as it reaches its maximum height?
(b) What is the kinetic energy of the shot at its maximum height?
(c) What is the kinetic energy of the shot just as it strikes the ground?
(d) Apply the law of conservation of energy to determine the final speed of the shot.
To answer these questions, we need to understand the concepts of mechanical energy, kinetic energy, and the law of conservation of energy.
(a) The total mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE). At its maximum height, the shot has no kinetic energy, so its total mechanical energy is equal to its potential energy.
To calculate the potential energy, we can use the formula PE = mgh, where m is the mass of the shot, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the ground.
PE = (7.26 kg)(9.8 m/s^2)(8.0 m) = 568.416 J
Therefore, the total mechanical energy of the shot as it reaches its maximum height is 568.416 Joules.
(b) At its maximum height, the shot has no kinetic energy. Therefore, the kinetic energy of the shot at its maximum height is zero.
(c) The kinetic energy of the shot just as it strikes the ground can be calculated using the formula KE = 1/2mv^2, where m is the mass of the shot and v is the velocity of the shot.
To find the velocity just before the shot reaches the ground, we can use the principle of conservation of energy. Since there is no air resistance, the total mechanical energy of the shot is conserved throughout its motion. Therefore, the total mechanical energy at the maximum height is equal to the total mechanical energy just before it strikes the ground.
The total mechanical energy at the maximum height is 568.416 J (as found in part a).
The total mechanical energy just before it hits the ground is equal to the sum of its potential energy at that height and its kinetic energy at that velocity.
Potential energy just before it hits the ground = mgh, where h is the height above the ground at that moment. Since it falls from 8.0 m, h = 8.0 m - 2.0 m = 6.0 m.
Potential energy just before it hits the ground = (7.26 kg)(9.8 m/s^2)(6.0 m) = 426.672 J
Therefore, the kinetic energy of the shot just as it strikes the ground is equal to the total mechanical energy minus the potential energy just before hitting the ground:
KE = Total Mechanical Energy - Potential Energy
= 568.416 J - 426.672 J
= 141.744 J
Therefore, the kinetic energy of the shot just as it strikes the ground is 141.744 Joules.
(d) The final speed of the shot can be found using the principle of conservation of energy. Total mechanical energy is conserved, so the total mechanical energy at the maximum height is equal to the total mechanical energy just before it strikes the ground.
The total mechanical energy at the maximum height is 568.416 J.
Let's assume the final speed of the shot is v_f.
Potential energy just before it hits the ground is (7.26 kg)(9.8 m/s^2)(6.0 m) = 426.672 J.
Kinetic energy just before it hits the ground is 141.744 J (as found in part c).
Therefore, the total mechanical energy just before it strikes the ground is:
Total Mechanical Energy = Potential Energy + Kinetic Energy
= 426.672 J + 141.744 J
= 568.416 J
Since total mechanical energy is conserved, we have:
Total Mechanical Energy at maximum height = Total Mechanical Energy just before it strikes the ground
568.416 J = (1/2)(7.26 kg)(v_f)^2 + (7.26 kg)(9.8 m/s^2)(2.0 m)
Rearranging the equation, we can solve for v_f:
(1/2)(7.26 kg)(v_f)^2 = 568.416 J - (7.26 kg)(9.8 m/s^2)(2.0 m)
Simplifying the equation:
(1/2)(7.26 kg)(v_f)^2 = 438.144 J
Multiply through by 2:
(7.26 kg)(v_f)^2 = 876.288 J
Divide through by (7.26 kg):
(v_f)^2 = 876.288 J / (7.26 kg)
Taking the square root of both sides:
v_f = sqrt(876.288 J / (7.26 kg))
Calculating the value:
v_f = 11.32 m/s (approximately)
Therefore, the final speed of the shot is approximately 11.32 m/s.