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Physics using the idea of conservation of energy

Cbarker1

Active member
Jan 8, 2013
255
Dear Everybody

There is a ball which has mass of 11 grams is thrown straight up with an initial speed of 4 m/s. Assume the air drag is negligible.

a. What is the kinetic energy of the ball (in J) as it leaves the hand? b. How much work is done by the gravitational force during the ball's rise to its zenith? c. What's the change in the gravitational potential energy of the ball during the ball's rise to its zenith? d. If the grav. force is taken zero at the pt. where it leaves your hands, what's the grav. pot. energy when it reaches the max. height? e. If the grav. force is taken zero at the pt. where it leaves your hands, what would the grav. pot. energy when it reaches the max. height? f. What is the max height the ball reaches?


The work:

The known values
mass is 11 grams
The initial velocity is 4m/s
The final velocity is 0
the value of gravity is 9.81m/s*s

a. I need to convert grams to kilograms. 11 grams to .011 kg. Then use kinetic energy which $K=(mv^2)/2$.
the answer is .088J.
 
Last edited:

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Yes, kinetic energy is "(1/2)mv^2". If you want the energy in Joules, you want the mass in kg and speed in m/s.

11 g grams is 11/1000= 0.011 kg. At the moment the ball is thrown up v is 4 m/s so the kinetic energy is (1/2)(0.011)(4^2)= 0.088 J


At the "zenith", the speed is 0 so there is no kinetic energy. "Using the idea of conservation of energy", that means that the increase in potential energy must be that same 0.088 J. The "work done" on the ball by the force of gravity is -0.088 J because that is how much its kinetic energy decreased. You say "If the grav. force is taken zero at the pt. where it leaves your hands" but you must mean "if the grav. potential energy is taken zero"- the force is always -mg, not 0. If the potential energy is, to start, 0 an increases by 0.088 J, as above, then it has increased to 0.088 J.

Your "e" is exactly the same as "d"??? Did you miscopy?

To find the maximum height, set mh= 0.088. h= 0.088/0.011= 1 m. That will be, of course, 1 meter above the point at which it is released.