Question

An object initially at an elevation of 5 m relative to Earth’s surface with a velocity of 50 m/s is acted on by an applied force R and moves along a path. Its final elevation is 20 m and its velocity is 100 m/s. The acceleration of gravity is 9.81 m/s2 . Determine the work done on the object by the applied force, in kJ

Answer 1

• 
  `(Work)/(mass)=   \ 3.897 (kJ)/(kg)`

Step-by-step explanation:

The work made by the force its equal to the change of mechanical energy

`Work = \Delta E_m`

The mechanical energy is given by:

`E_m = E_k + E_p`

where
`E_k` is the kinetic energy and
`E_p` the potential energy.

We know that, the kinetic energy is

`E_k = (1)/(2) \ m \ v^2`

and the potential energy

`E_p =m \ g  \ h`

We can put h=0 at Earth's surface, taking this in consideration, the initial Energy is:

`E_(m_0) = (1)/(2) \ m \ (50 (m)/(s))^2 + m \ 9.81 (m)/(s^2) \ 5 \ m`

`E_(m_0) = (1)/(2) \ m \ 2,500 (m^2)/(s^2) + m 49.05 (m^2)/(s^2)`

`E_(m_0) = \ m \ 1,250 (m^2)/(s^2) + m \ 49.05 (m^2)/(s^2)`

`E_(m_0) = \ m \ 1,299.05 (m^2)/(s^2)`

the final Energy is:

`E_(m_f) = (1)/(2) \ m \ (100 (m)/(s))^2 + m \ 9.81 (m)/(s^2) \ 20 \ m`

`E_(m_f) = (1)/(2) \ m \ 10,000 (m^2)/(s^2) + m  196.2 (m^2)/(s^2)`

`E_(m_f) = \ m \ 5,000 (m^2)/(s^2) + m \ 196.2 (m^2)/(s^2)`

`E_(m_f) = \ m \ 5,196.2 (m^2)/(s^2)`

The work will be:

`Work = \Delta E_m`

`Work =  \ m \ 5,196.2 (m^2)/(s^2) - \ m \ 1,299.05 (m^2)/(s^2)`

`Work =  \ m \ 3,897.15 (J)/(kg)`

As we don't know the mass, we can take the work over mass:

`(Work)/(mass)=   \ 3.897 (kJ)/(kg)`