1. A 55.0 kg pole vaulter running at 11.0 m/s vaults over the bar. Assuming that the vaulter's horizontal component of velocity over the bar is 1.0 m/s and disregarding air resistance, how high was the jump (include units)?
2. A 71 kg person rolls on a 2 kg skateboard with a velocity of 6 m/s. The skateboarder runs into a large rubber band with a spring constant of 80 N/m. How far will the cord stretch till the skateboarder stops moving forward (include units)?
kinetic energy coming to the jump = (1/2) m(121)
kinetic energy at the top = (1/2) m (1)
loss of Ke = (1/2) m (120)
= gain of potential = m g h
(note that the mass is irrelevant, big person goes as high as small one if same speeds)
so
(1/2)(120) = 9.81 h
solve for h
ke = (1/2) m v^2
goes until that energy is stored in spring
so
(1/2) m v^2 = (1/2) k x^2
To find the answers to these questions, we can use some basic principles of physics.
1. To determine the height of the jump, we can use the law of conservation of mechanical energy. We can calculate the potential energy gained by the vaulter and equate it to the initial kinetic energy. The formula for potential energy (PE) is given by PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height. The initial kinetic energy (KE) can be calculated using the formula KE = 0.5mv^2, where v is the velocity.
First, we can find the initial kinetic energy:
KE = 0.5 * 55.0 kg * (11.0 m/s)^2
KE = 3332.5 J
Since the horizontal component of velocity over the bar is 1.0 m/s, it does not contribute to the potential energy. Therefore, we subtract the kinetic energy due to the horizontal velocity:
KE = 3332.5 J - 0.5 * 55.0 kg * (1.0 m/s)^2
KE = 3331.5 J
Now, equating the potential energy and the remaining kinetic energy:
mgh = KE
h = KE / (mg)
h = (3331.5 J) / (55.0 kg * 9.8 m/s^2)
h ≈ 6.0 m
Therefore, the height of the jump is approximately 6.0 meters.
2. To determine how far the cord will stretch till the skateboarder stops moving forward, we can use the principle of work-energy. The work done by the cord will be equal to the change in kinetic energy of the skateboarder. We can calculate this using the formula W = ΔKE = 0.5mvf^2 - 0.5mvi^2, where m is the mass, vf is the final velocity, and vi is the initial velocity.
The initial kinetic energy of the system is:
KEi = 0.5 * (71 kg + 2 kg) * (6 m/s)^2
KEi = 1269 J
The final kinetic energy is zero since the skateboarder stops moving forward. So, we have:
W = ΔKE = -1269 J
The work done by the cord can be calculated as the force applied by the cord multiplied by the distance stretched (d):
W = Fd
The force applied by the cord can be found using Hooke's Law:
F = kx, where k is the spring constant and x is the displacement from the equilibrium position.
Rearranging the equation for work done by the cord, we have:
Fd = -1269 J
Substituting the force in terms of displacement:
kx * d = -1269 J
Simplifying and rearranging the equation, we get:
d = -1269 J / (kx)
Plugging in the values, we have:
d = -1269 J / (80 N/m * x)
To find the distance stretched (d), we need to know the specific displacement (x) of the cord. Without knowing this value, we cannot calculate the exact distance.