You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it is traveling at a speed of 25.0m/s upward.
A) Use the work-energy theorem to find its speed just as it left the ground. What is it?
B) Use the work-energy theorem to find its maximum height. What is it?
See previous post: Mon,3-17-14,8:03 PM.
A) To find the speed of the rock just as it left the ground using the work-energy theorem, we need to consider the initial and final kinetic energies.
The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy of the rock when it's at ground level is zero since its speed is zero.
The final kinetic energy of the rock when it's at a height of 14.8m above the ground is given by:
Kf = (1/2) * m * vf^2
where m is the mass of the rock and vf is the final velocity (25.0 m/s upward).
Since the rock is thrown vertically, the only force acting on it is gravity, which does negative work on the rock. This means the work done by gravity is equal to the negative change in kinetic energy.
The work done by gravity is given by:
W = -m * g * h
where g is the acceleration due to gravity (9.8 m/s^2) and h is the height (14.8m).
The change in kinetic energy is given by:
ΔK = Kf - Ki
= (1/2) * m * vf^2 - 0
= (1/2) * m * (25.0 m/s)^2
Since the work done by gravity is equal to the negative change in kinetic energy, we have:
-m * g * h = -ΔK
m * g * h = ΔK
Plugging in the values:
(1/2) * m * (25.0 m/s)^2 = m * (9.8 m/s^2) * (14.8 m)
We can cancel out the mass:
(1/2) * (25.0 m/s)^2 = 9.8 m/s^2 * 14.8 m
Solving for vf^2:
vf^2 = (2 * 9.8 m/s^2 * 14.8 m) / (25.0 m/s)^2
Taking the square root of both sides:
vf = √((2 * 9.8 m/s^2 * 14.8 m) / (25.0 m/s)^2)
Evaluating this expression will give us the speed of the rock just as it left the ground.
B) To find the maximum height reached by the rock using the work-energy theorem, we can consider the initial and final kinetic energies again.
At the maximum height, the final kinetic energy is zero since the rock comes to a stop.
Using the work-energy theorem, we have:
W = -m * g * h_max
The change in kinetic energy is:
ΔK = Kf - Ki
= 0 - (1/2) * m * vf^2
Since the work done by gravity is equal to the negative change in kinetic energy, we have:
-m * g * h_max = -ΔK
m * g * h_max = (1/2) * m * vf^2
Canceling out the mass, we have:
g * h_max = (1/2) * vf^2
Solving for h_max, we get:
h_max = (1/2) * vf^2 / g
Plugging in the values:
h_max = (1/2) * (25.0 m/s)^2 / 9.8 m/s^2
Evaluating this expression will give us the maximum height reached by the rock.
To solve this problem using the work-energy theorem, we need to consider the work done on the rock, the change in kinetic energy, and the gravitational potential energy.
A) To find the speed just as the rock left the ground, we can equate the initial kinetic energy (KE) to zero and the final kinetic energy to the change in gravitational potential energy (ΔPE):
KE_initial + ΔPE = KE_final
The initial kinetic energy is zero because the rock starts from rest, so the equation becomes:
0 + ΔPE = KE_final
The change in gravitational potential energy is given by:
ΔPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
We are given the height h = 14.8 m and the final speed KE_final = 25.0 m/s. We need to find the mass (m) to calculate the change in gravitational potential energy.
Now, we can rearrange the equation and solve for the mass (m):
ΔPE = mgh → m = ΔPE / (gh)
The change in gravitational potential energy is:
ΔPE = KE_final = (1/2)mv^2, where v is the final velocity
Substituting the given values:
25.0^2 = (1/2) * m * 14.8 * 9.8
Now, solve for the mass (m):
m = (25.0^2) / (0.5 * 14.8 * 9.8)
Once you have found the mass, you can substitute it back into the equation for ΔPE to find the value of ΔPE (change in gravitational potential energy) at the height of 14.8m above the ground.
B) To find the maximum height, we can equate the initial kinetic energy (KE) to the change in gravitational potential energy (ΔPE):
KE_initial = ΔPE
The initial kinetic energy is given by:
KE_initial = (1/2)mv^2, where v is the velocity just as the rock left the ground.
Using the mass calculated in part A, and the velocity calculated in part A (v), we can substitute these values into the equation to find the initial kinetic energy.
Then, we can equate this to the change in gravitational potential energy to solve for the maximum height.
ΔPE = mgh
Substituting the values from part A, we can now solve for the height.