PHYSICS question involving work?

Question

A 60 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 3.5 m/s. (Ignore small changes in gravitational potential energy.)
a)How fast is he going as he lands on the trampoline, 3.0 m below? (This is 8.43 m/s)
(b) If the trampoline behaves like a spring with spring stiffness constant 5.2 104 N/m, how far does he depress it?

Answer 1

The easiest way to do this is using energy.

We can find the amount of kinetic energy the person has right before she hits the trampoline, and say that that energy is the same amount of energy stored by the trampoline when it is fully depressed plus the potential energy it has from the distance away from the trampoline.

The equation for kinetic energy is (1/2)mv^2
The equation for elastic potential energy is (1/2)kX^2
The equation for potential energy due to gravity is mgX

m is mass, v is velocity, k is spring constant, X is distance of depression

So we setup our equation:

(1/2)(60)(8.43)^2 = (1/2)(5.2104)(X)^2 + (60)(9.81)(x)



And we solve for X.

X comes out to be 3.569 meters

Answer 2

The work done by non-conservative forces equals the change in energy.

Wnc = Ef - Ei

The number 1 non-conservative force? Friction/air resistance!

That's not here, so the work done by nc forces is zero!

0 = Ef - Ei, or

Ef = Ei

KEf + PEf = KEi + PEi

½ m vf² + m g hf = ½ m vi² + m g hi

Cancel out all of the masses!!!

½ vf² + g hf = ½ vi² + g hi

0.5 * vf² + 9.8 * (-3) = 0.5 * (3.5)²

vf = 8.429 m/s

So you're answer for a is correct!

b) Realize that springs have potential energy (it takes work to compress it!).

PE(springs) = ½ k * x²
PE(gravity) = m * g * h

where k = spring constant (spring stiffness)
x = distance depressed

I'm assuming you meant k = 5.2 * 10^4

Since there are still no nc forces,

Ef = Ei

KEf + PEf = KEi + PEi

½ m vf² + ½ k xf² + m g hf = ½ m vi² + ½ k xi² + m g hi

½ * (5.2 * 10^4) * x² + 60 * 9.8 * x = ½ * 60 * 8.43²

26000 x² + 588 x= 2131.95

x = -0.2978 m, or 0.298 m downward

Good luck!

Answer 3

In a loss-less (no friction) model of the physical world, energy conversion from potential to kinetic is an easy way to predict motion of objects.

In this case the artist starts with kinetic energy
.5*m*3.5^2
and gains additional kinetic energy through a 3 m drop
m*g*3

which gets converted to potential energy in the springs of the trampoline.

I will use 10 as g

The kinetic energy when he lands is

.5*60*3.5^2+60*10*3
=2167.5 J

A spring stores energy at a rate of
.5*k*x^2

Ignoring the additional potential energy loss

x=sqrt(2*2167.5/52000)
=.3 m

If you add the gravitational potential energy loss to the spring energy, just make m*g*h be
m*g*3+x
plug into the energy conservation equation above and solve for x.

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