Question

A person throws a 10kg rock straight up. If the person exerted 3500N of force on the rock over a distance

of 0.5m, how high does the rock move?
Pls help and can you tell me the energy conversions too!!

Answer 1

h = 17.83[m]

Step-by-step explanation:

By means of the working equation which is equal to the product of force by distance, we can find the energy printed at the launch.

`W=F*d`

where:

W = work [J]

F = force = 3500 [N]

d = distance = 0.5 [m]

Now replacing:

`W= 3500*0.5\\W=1750[J]`

Now, this same work is converted to kinetic energy, necessary for the body to move with an initial velocity. And by energy conservation, we can say that kinetic energy is transformed into potential energy. That is, the kinetic energy will be equal to the potential energy.

`W=E_(kin)=E_(pot)`

where:

Ekin = kinetic energy [J]

Epot = potential energy [J]

Therefore:

`1750 = m*g*h\\1750=10*9.81*h\\h = 17.83[m]`