How much potential energy did the bowling ball have when you held it to your face?
PE = mgh
The mass of the bowling ball is 5.5 kg if you had the ball 1.7 m above the ground what was its potential energy
Assuming no energy was lost how fast was it going at the bottom of the swing (hint: sub PE from above in for KE and solve for velocity in the KE equation)?
The potential energy of the bowling ball when held 1.7 m above the ground can be calculated using the formula PE = mgh, where m is the mass (5.5 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (1.7 m).
PE = mgh
PE = (5.5 kg)(9.8 m/s^2)(1.7 m)
PE ≈ 91.37 J
The potential energy of the bowling ball when held to the face is approximately 91.37 joules.
To find the velocity of the ball at the bottom of the swing, assuming no energy was lost, we can use the law of conservation of energy.
The total energy at the top of the swing (potential energy) is the same as the total energy at the bottom of the swing (kinetic energy).
Potential energy at the top = Kinetic energy at the bottom
mgh = 1/2 mv^2
Substituting the values, we have:
(5.5 kg)(9.8 m/s^2)(1.7 m) = 1/2 (5.5 kg) v^2
Simplifying,
91.37 J = 1/2 (5.5 kg) v^2
Multiplying both sides by 2,
182.74 J = 5.5 kg v^2
Dividing both sides by 5.5 kg,
v^2 = 33.22 m^2/s^2
Taking the square root of both sides,
v ≈ 5.76 m/s
Therefore, the speed of the bowling ball at the bottom of the swing, assuming no energy was lost, is approximately 5.76 m/s.