Given that coal used by electric power plants has a heating value of 27.5 million btus metric ton (25 million btus per ton), det
1 answer:
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1) The potential energy is the most at the highest position and the least at the equilibrium position
2) The kinetic energy is the most at the equilibrium position and the least at the highest position
Explanation:
1)
The potential energy of an object is the energy possessed by the object due to its position in a gravitational field; mathematically, it is given by
where
m is the mass of the object
g is the strength of the gravitational field
h is the height of the object relative to the ground
For the pendulum in this problem, m is the mass of the bob, and h is the height of the above relative to the ground. We see from the formula that the potential energy is directly proportional to the height:
This means that:
- The potential energy is the most when the bob is at the highest position
- The potential energy is the least when the bob is at the equilibrium position, which is the lowest position
2)
We can solve this part by applying the law of conservation of energy: in fact, the total mechanical energy of the pendulum (sum of potential and kinetic energy) is constant at any time during the motion,
where KE is the kinetic energy.
From the equation above, we observe that:
- When PE is maximum, KE must be at minimum
- When PE is minimum, KE must be maximum
Therefore, this implies that:
- The kinetic energy is the most when the potential energy is the least, i.e. at the equilibrium position
- The kinetic energy is the least when the potential energy is the most, i.e. at the highest position
Learn more about kinetic and potential energy:
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Answer:
a.
b. must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c. is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,
- speed of lagging car,
- distance between the cars,
- deceleration of the leading car after braking,
a.
Time taken by the car to stop:
where:
, final velocity after braking
time taken
b.
using the eq. of motion for the given condition:
where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:
is the time taken to stop after braking