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

Answer:

• $\frac{Work}{mass}=   \ 3.897 \frac{kJ}{kg}$

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

$E_k = \frac{1}{2} \ m \ v^2$

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:

$E_{m_0} = \frac{1}{2} \ m \ (50 \frac{m}{s})^2 + m \ 9.81 \frac{m}{s^2} \ 5 \ m$

$E_{m_0} = \frac{1}{2} \ m \ 2,500 \frac{m^2}{s^2} + m 49.05 \frac{m^2}{s^2}$

$E_{m_0} = \ m \ 1,250 \frac{m^2}{s^2} + m \ 49.05 \frac{m^2}{s^2}$

$E_{m_0} = \ m \ 1,299.05 \frac{m^2}{s^2}  $

the final Energy is:

$E_{m_f} = \frac{1}{2} \ m \ (100 \frac{m}{s})^2 + m \ 9.81 \frac{m}{s^2} \ 20 \ m$

$E_{m_f} = \frac{1}{2} \ m \ 10,000 \frac{m^2}{s^2} + m  196.2 \frac{m^2}{s^2}$

$E_{m_f} = \ m \ 5,000 \frac{m^2}{s^2} + m \ 196.2 \frac{m^2}{s^2}$

$E_{m_f} = \ m \ 5,196.2 \frac{m^2}{s^2}  $

The work will be:

$Work = \Delta E_m$

$Work =  \ m \ 5,196.2 \frac{m^2}{s^2} - \ m \ 1,299.05 \frac{m^2}{s^2}$

$Work =  \ m \ 3,897.15 \frac{J}{kg}$

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

$\frac{Work}{mass}=   \ 3.897 \frac{kJ}{kg}$