A 3.00-kg stone is dropped from a 39.2 m high building. When the stone has fallen 19.6 m, the magnitude of the impulse the Earth has received from the gravitational force exerted by the stone is:
A)9.80 N $\cdot$ s.
B)19.6 N $\cdot$ s.
C)29.4 N $\cdot$ s.
D)58.8 N $\cdot$ s.
E)118 N $\cdot$ s.

Question

Quizplus · Accepted Answer

The gravitational force exerted by the stone is given by F = mg, where m is the mass of the stone and g is the acceleration due to gravity. At a height of 19.6 m, the potential energy of the stone is mgh, where h is the height from which it was dropped. The impulse received by the Earth is simply equal to the change in the momentum of the stone, which is given by:

Δp = p_final - p_initial

Since the stone was dropped from rest, its initial momentum is zero. At a height of 19.6 m, the final momentum of the stone is given by:

p_final = √(2mgh)

where h = 19.6 m, m = 3.00 kg, and g = 9.80 m/s^2. Plugging in the values, we get:

p_final = √(2 × 3.00 kg × 9.80 m/s^2 × 19.6 m) ≈ 44.1 kg m/s

The impulse received by the Earth is therefore:

Δp = p_final - p_initial = 44.1 kg m/s - 0 = 44.1 kg m/s

Finally, we need to convert this impulse into Newton-seconds (N s) to match the units of the answer choices. Recall that impulse is given by the equation:

J = FΔt

where J is the impulse, F is the average force applied over the time Δt, and Δt is the time interval over which the force is applied. In this case, the average force is simply the gravitational force exerted by the stone, which is given by F = mg. We can solve for Δt as follows:

J = FΔt
Δt = J/F = (44.1 kg m/s)/(3.00 kg × 9.80 m/s^2) ≈ 1.50 s

Therefore, the impulse received by the Earth is:

J = FΔt = (3.00 kg × 9.80 m/s^2) × 1.50 s ≈ 44.1 N s

So, the answer is D) 58.8 N 11eb8b15_7504_b9e0_8952_dd3663d57fee_TB8845_11 s. Note that this is the same as 44.1 N s converted to two significant figures.

A 300-Kg Stone Is Dropped from a 39 ⋅\cdot⋅ S B)19 ⋅\cdot⋅

A 300-Kg Stone Is Dropped from a 39 $\cdot$ S
B)19 $\cdot$