Question

A 300-kg piano being held by a crane is accidentally dropped from a height of 15 meters. a. What is the speed of the piano just before it hits the ground? b. How much energy is dissipated when the piano hits the ground? c. What are some indicators of this release of energy that could be observed during and shortly after the piano hits the ground? d. Suppose that a parachute is used to slow the fall of the piano. i. How would the amount of kinetic energy dissipated by the ground change? Explain. ii. How would the total amount of energy dissipated change? Explain.

Answer 1

Answer:

a) 17.16m/s

b) 44,145J

c) Sound the piano makes when hitting the ground, vibration of the ground, heat.

d) i) It's smaller due to the energy dissipated by the friction between air and the parachute.

ii) It stays the same, the only difference is that the dissipated energy is distributed between air resistance and the kinetic energy dissipated by the ground whent he piano hits it.

Explanation:

a)

In order to solve this problem we must start by doing a drawing of the situation, which will help us visualize the problem better. (See attached picture).

So, in this problem we can ignore air resistance so we can say that the energy is conserved, this is the total initial energy is the same as the total final energy, so we get that:

$U_{0}+K_{0}=U_{f}+K_{f}$

When the piano is released it has an initial speed of zero, so the initial kinetic energy is zero. When the piano hits the ground it will have a height of 0m, so the final potential energy is zero as well. This will simplify our equation:

$U_{0}=K_{f}$

We know that potential energy is given by the formula:

U=mgh

and kinetic energy is given by the formula:

$K=\frac{1}{2}mv^{2}$

which can be substituted in our equation:

$mgh=\frac{1}{2}mv^{2}$

we can divide both sides of the equation into the mass of the piano, so we get:

$gh=\frac{1}{2}v^{2}$

which can be solved for the final velocity which yields:

$v=\sqrt{2gh}$

we can now substitute the data provided by the problem so we get:

$v=\sqrt{2(9.81m/s^{2})(15m)}$

which yields:

v=17.16m/s

b)

Since energy is conserved, this means that the total dissipated energy will be the same as the potential energy, so we get that:

E=mgh

so

$E=(300kg)(9.81m/s^{2})(15m)$

which yields:

E=44,145J

c)

When the piano hits the ground, the kinetic energy it had will be transformed to other types of energy, mostly vibration and heat. The vibration will turn to sound due to the movement of air created by the piano itself and the ground. And heat is created by the friction between the molecules created by the vibrations and the collition itself. So some of the indicators of this release of energy could be:

-Sound

-Vibration

-Heat.

d)

i) The amount of inetic energy dissipated would decrease due to the friction between air and the parachute. Since air is resisting the movement of the piano, this will translate into a loss of energy, if we did an energy balance we would get that:

$U_{0}=K_{f}+E_{p}$

The total amount of energy is conserved but it will be distributed between the energy lost due to air resistance and the kinetic energy the piano has at the time it hits the ground.

ii) So the total amount of energy dissipated remains the same, the only difference is that it will be distributed between air resistance and the kinetic energy of the piano.