a hammer falls off a rooftop and strikes the ground with a certain kinetic energy. if it fell from a roof twice as tall, how would its kinetic energy compare? explain
mgh = KE
mg2h = 2KE
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To explain how the kinetic energy of the falling hammer would compare if it fell from a rooftop twice as tall, we need to understand the relationship between potential energy and kinetic energy.
When an object is lifted to a certain height, it gains potential energy due to its position relative to the ground. This potential energy can then be converted into kinetic energy as the object falls, according to the law of conservation of energy.
The potential energy (PE) of an object at a given height is given by the formula:
PE = m * g * h
Where:
m = mass of the object
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the object
Now, let's compare the two scenarios:
Scenario 1: The hammer falls from a certain height (h) and strikes the ground with a certain kinetic energy.
Scenario 2: The same hammer falls from a rooftop twice as tall (2h).
In both scenarios, the mass of the hammer (m) remains the same, as it is referring to the same object.
In Scenario 2, the height (h) is doubled, so the potential energy (PE) of the hammer at the rooftop is given by:
PE2 = m * g * (2h)
Since the kinetic energy (KE) is derived from the potential energy, we can equate the two as follows:
KE1 = PE1 = m * g * h
KE2 = PE2 = m * g * (2h)
Therefore, if the hammer fell from a rooftop twice as tall, its kinetic energy when striking the ground would also be twice as much.