# A baby elephant with a mass of 200 kg bounces on a trampoline as shown below. The surface of the...

## Question:

A baby elephant with a mass of 200 kg bounces on a trampoline as shown below. The surface of the trampoline is 1.0 m above the ground. The trampoline compresses 0.76 m and the elephant exits with a speed of 8 m/s.

A) How high above the ground does the elephant travel?

B) What is the stiffness of the trampoline?

## Conservation Of Energy:

When a particle moves in a conservative force field its total mechanical energy is conserved. The mechanical energy is constituted of two parts. One is the kinetic energy that arises from the state of motion of the particle. The other is the potential energy that arises on account of the position of the particle in an external force field. The mathematical expression for the potential energy will depend on the exact nature of the force field. Near the surface of the earth the gravitational potential energy is a linear function of the position. On the other hand, the elastic potential energy due to a spring is a quadratic function of the position.

## Answer and Explanation:

The baby elephant has a mass of {eq}\displaystyle {m=200\ kg} {/eq}.

The trampoline is at a height of {eq}\displaystyle {H=1\ m} {/eq} from the floor.

The baby elephant bounces on the trampoline compressing it through a distance of {eq}\displaystyle {a=0.76\ m} {/eq}.

The elephant will lose contact with the trampoline at the instant the trampoline gets back to its original configuration 1 m above the floor.

The total energy of the elephant-trampoline system is conserved during the entire motion.

A. The elephant exits the trampoline with a speed of {eq}\displaystyle {v=8\ m/s} {/eq}.

Therefore its kinetic energy at the time of exit is {eq}\displaystyle {KE=\frac{1}{2}mv^2=\dfrac 1 2 \times 200\times 8^2=6400\ J} {/eq}.

The elephant will continue its upward journey at the expense of its kinetic energy. At the maximum possible height, say {eq}\displaystyle {h } {/eq} from the trampoline, the kinetic energy is completely converted to its gravitational potential energy.

Thus we have the condition : {eq}\displaystyle {mgh=KE } {/eq}

{eq}\displaystyle {\text{or} \quad h=\frac{KE}{mg}=\frac{6400}{200\times 9.8}=3.27 \ m } {/eq}

Since the trampoline is 1.0 m above the ground, the elephant will rise to a height of {eq}\displaystyle { h+1.0=3.27+1.0=4.27 \ m } {/eq} from the ground.

B. Say the trampoline has a stiffness constant {eq}\displaystyle { k } {/eq}. At maximum compression, the elastic potential energy of the trampoline is {eq}\displaystyle { E=\frac{1}{2}ka^2=\dfrac 1 2 k\times 0.76^2\ J=0.2888 k\ J } {/eq}

As the elephant starts to rise, the elastic potential energy of the trampoline is converted to the gravitational potential energy and the kinetic energy of the elephant.

The elephant rises through 0.76 m before it exits the trampoline. Then the increase in the gravitational potential energy is {eq}\displaystyle {\Delta U= mga=200\times 9.8\times 0.76=1489.6\ J} {/eq}.

Again we have found in the previous section that the elephant's kinetic energy at this point is {eq}\displaystyle { KE=6400 \ J } {/eq}

Then, using the conservation of energy,

{eq}\displaystyle {E= \Delta U+KE} {/eq}

That is,

{eq}\displaystyle { 0.2888k=1489.6+6400} {/eq}

That is,

{eq}\displaystyle { k=27319\ N/m} {/eq}

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