Question

A 700-N man jumps out of a window into a fire net 10 m below. The net stretches 2 m before

bringing the man to rest and tossing him back into the air. The maximum potential energy of

the net, compared to its unstretched potential energy, is:

Answer 1

The maximum potential energy of the net, compared to its unstretched potential energy, is:
`E_(mn) -E_(un) = 8400J`

Step-by-step explanation:

From the question we are told that

The weight of the man is
`W_m = 700 \ N`

The distance of the window to the net is
`d = 10 \ m`

The distance stretched by net is
`D = 2 \ m`

Generally from the of conservation energy , the total energy is conserved

This implies that

`E_m__(i) + E_(un) = E_m__(f)} + E_(mn)`

Where
`E_m__(i)}` is the initial potential energy of the , man which is mathematically evaluated as

`E_m__(i)} = W_m * d`

Substituting values

`E_m__(i)} = 700 *10`

`E_m__(i)} = 7000 \ J`

And
`E_(un)` is the unstretched potential energy of the net

And
`E_m__(f)}` is the final potential energy of the man which is mathematically evaluated as

`E_m__(f)} = - W_m * D`

Substituting values

`E_m__(f)} = - 700 * 2`

`E_m__(f)} = - 1400`

The negative show that the direction of the man is against the direction of gravitational pull

And
`E_(mn)` is the maximum potential energy of the net when stretched

So the above equation becomes

`7000 + E_(un) = - 1400 + E_(mn)`

So

`E_(mn) -E_(un) = 7000 + 1400`

`E_(mn) -E_(un) = 8400J`