Inquiry into Physics
8th Edition
ISBN: 9781337515863
Author: Ostdiek
Publisher: Cengage
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Chapter 3, Problem 8C
To determine
To prove:
The work done by a force is equal to the kinetic energy of the object.
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Since the initial potential energy and final kinetic energy are zero, our equation now becomes
EP,f = EK,i.
We can then substitute the formula for kinetic energy,
EK = 1/2mv2,
and the formula for gravitational potential energy,
EP = mgh,
mghf = 1/2mvi2.
Now it's just a matter of doing the algebra, solving for the final height
hf,
and substituting values to find
hf.
Notice that the mass m divides out of both sides of the equation, so the value of the mass is not needed to find the final height.
Calculate the maximum height of the ball in meters.
hf = __________ m
From a building which is 60 meters height, a ball was thrown horizontally with kinetics energy of 45 Joules. If the ball weighs 300 grams, compute for the ball’s kinetic energy when it touches the ground. Give your answer in Joules. (KE = 1/2mV1^2)
Suppose that you want to build a tower out of bricks. Suppose that each brick has a mass of 0.5 kg and is 0.1 m tall. You want the tower to be 20 bricks high, and we want to calculate how much energy is required to build it. The physics fact that you need is this: the energy, E (in Joules) required to lift a mass m (in kg) a height h (in m) is E = mgh (where g is the acceleration due to gravity - for this problem, you can take g to be 10 m/s2.)
1. We will assume that no energy is required to place the first brick in the tower, since it is already on the ground. What about the second brick? How high must if be lifted to be placed on top of the first brick? How much energy does that require? What about the third brick? The tenth brick?
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Inquiry into Physics
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