A pendulum is used in a large clock. The pendulum has a mass of 2 kg. If the pendulum is moving at a speed of 2.9 m/s when it re
2 answers:
This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
Solving for h gives us:
It doesn't depend on mass!
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
Solving for h gives us:
It doesn't depend on mass!
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Answer:
a) 6 times farther. b) 6 times longer.
Explanation:
Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:
vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s
In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:
vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s
Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:
If the total displacement in the vertical direction is 0 (which means that the time if the total time of flight), we can solve for t, as follows:
On earth, this time could be calculated in the same way:
As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:
Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m
On earth, the distance travelled had been as follows:
Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m
⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00
b) As we have just found, the time of flight on the moon and on the earth are as follows:
tmoon = 11. 9 s
tearth = 1.98 s
⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0