You throw a 3.00-N rock vertically into the air from ground level. You observe that when it
is 15.0 m above the ground, it is traveling at 25.0 m/s upward. Use the work–energy theorem
to find
a. the rock’s speed just as it left the ground
(1/2)mv^2 + mgh = (1/2) mv^2+ mgh, m cancels, does not matter
(1/2) v^2 + 0 = (1/2) * 625 + 9.81*15 solve for v
To find the rock's speed just as it left the ground using the work-energy theorem, we can use the principle of conservation of mechanical energy.
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done on the rock is equal to the change in its kinetic energy from the moment it was thrown until it reached a height of 15.0 m.
Since the rock was thrown vertically, we can assume that there is no horizontal motion, and therefore the only energy involved is gravitational potential energy and kinetic energy.
We can express the work done on the rock as the sum of the initial kinetic energy (K1) and the initial potential energy (PE1), and the final kinetic energy (K2) and final potential energy (PE2) at a height of 15.0 m.
When the rock is at ground level, its height (h) is zero, so the gravitational potential energy (PE) is zero. Therefore, PE1 = 0.
At a height of 15.0 m, the potential energy is PE2 = mgh, where m is the mass of the rock, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (15.0 m).
Since the rock is at its maximum height, its final velocity is zero (as it reaches the highest point in its trajectory). Therefore, K2 = 0.
The work done on the rock is equal to the change in kinetic energy, so we have:
Work = K2 - K1 = PE2 - PE1
Substituting the values we have:
0 - K1 = mgh - 0
-K1 = mgh
K1 = -mgh
Since we are interested in the magnitude of the initial kinetic energy (speed), we can ignore the negative sign.
Therefore, the magnitude of the initial kinetic energy (speed) of the rock just as it left the ground is given by:
|K1| = mgh
Using the given values, let's substitute them into the equation:
|K1| = (mass of the rock) * (gravitational acceleration) * (height)
|K1| = 3.00 N * (9.8 m/s^2) * (15.0 m)
Solving this equation will give us the magnitude of the initial kinetic energy (speed) of the rock just as it left the ground.