Question

You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 16.0 m above the ground, it is traveling at 22.0 m/s upward. Use the work-energy theorem to find the rock's speed just as it left the ground.

Answer 1

Using the work-energy theorem, the speed of the rock just as it left the ground can be calculated by equating the gravitational potential energy lost by the rock to its change in kinetic energy. After finding the rock's mass and gravitational potential energy, we can use the formula for kinetic energy to find its speed.

Step-by-step explanation:

To find the rock's speed just as it left the ground, we can use the work-energy theorem. The work-energy theorem states that the change in the object's kinetic energy is equal to the net work done on the object. In this case, the net work done on the rock is equal to its change in kinetic energy, which is equal to its final kinetic energy minus its initial kinetic energy.

First, let's find the rock's final kinetic energy when it is 16.0 m above the ground. The gravitational potential energy lost by the rock is equal to the work done on it by gravity, since there is no air resistance. Using the formula for gravitational potential energy, we have:

ΔPE = PEf - PEi = mgh

Where m is the mass of the rock, g is the acceleration due to gravity, and h is the height above the ground. Since we know the rock's weight is 3.00 N, we can use this information to find its mass:

F = mg → m = F/g = 3.00 N/9.8 m/s^2 ≈ 0.306 kg

Plugging in the values, we get:

ΔPE = mgh = 0.306 kg × 9.8 m/s^2 × 16.0 m = 47.04 J

Next, we can use the work-energy theorem to find the rock's final kinetic energy:

ΔKE = KEf - KEi

Since the rock is at rest on the ground, its initial kinetic energy is zero. Therefore, we have:

ΔKE = KEf - 0 = KEf

Now, we can set the gravitational potential energy equal to the change in kinetic energy:

ΔPE = ΔKE

47.04 J = KEf

Finally, we can find the speed of the rock just as it left the ground by using the formula for kinetic energy:

KE = 1/2mv^2

Plugging in the values, we get:

47.04 J = 1/2 × 0.306 kg × v^2

Solving for v, we find:

v^2 = 2(47.04 J)/(0.306 kg)

v^2 ≈ 308.24 m^2/s^2

v ≈ √(308.24 m^2/s^2) ≈ 17.57 m/s

Therefore, the rock's speed just as it left the ground is approximately 17.57 m/s.

Answer 2

The final speed of the stone as it lift the ground is 23.86 m/s.

Step-by-step explanation:

Given that,

Force acting on the rock, F = 3 N

Distance, d = 16 m

Initial speed of the stone, u = 22 m/s

We need to find the rock's speed just as it left the ground. It can be calculated using work energy theorem as :

`W=\Delta E\\\\W=(1)/(2)m(v^2-u^2)\\\\Fd=(1)/(2)m(v^2-u^2)\\\\v^2=(2Fd)/(m)+u^2\\\\v^2=(2mgd)/(m)+u^2\\\\v^2=2* 9.8* 16+(16)^2\\\\v=23.86\ m/s`

So, the final speed of the stone as it lift the ground is 23.86 m/s.