The Work-Energy Theorem

Work is the force times the distance covered by an object. When a force is applied to an object, the object moves to a distance in the direction of the applied force. To do work, energy is needed. Energy is defined as the capacity to do work. Thus, this only proves the relationship of work and energy. When there is energy, work can be done on an object. The energy possessed by the object is kinetic energy, since the work done causes an object's motion.

Consider an illustration below of the work-energy relationship.

 

When a force, \(\vec{F}\) is applied to an object with mass, \(m\), the object displaces to a distance, \(d\). The force times the distance covered gives the work done on the object. Since, the object is moving, it possesses kinetic energy. Thus, mathematically the relationship can be expressed as

\(W=KE\\ \vec{F} \times d=\frac 12m\Delta v^2\\ \vec{F} \times d=\frac 12m(v_f - v_i)^2\)

For the work done in an angle, the above equation can be written as

\(\vec{F} \times d\;\cos \theta=\frac 12m(v_f - v_i)^2\)


Example 1.

\(250-N\) force is applied to a loaded box with a mass of \(50\;kg\). How much work is done on the box if from rest, it is moved to a distance \(2.0\;m\). What is the object's velocity?

Given:

\(F=250\;N\\ m=50\;kg\\ d=2\;m\)


Solution 1.

Solve for work using \(W=\vec{F}\times d\).

\(W=(250\;N)(50\;kg)=500\;J\)


Solution 2.

\(W=KE \implies W=\frac 12 mv^2\) . To solve for the velocity of the object, we derive the formula for velocity as

\(W=\frac 12mv^2 \implies 2W=mv^2\\ \quad\quad\quad\quad\quad\implies v^2=\frac {2W}{m}\\ \quad\quad\quad\quad\quad\implies v=\sqrt \frac {2W}{m}\)

 

\(v=\sqrt \frac {2(500\;J)}{50\;kg}\\ \;\,=\sqrt \frac {1000\;J}{50\;kg}\\ \;\,=\sqrt {20\;m^2/s^2}\\ \;\,=4.47\;m/s\)

hence, the velocity of the object is \(4.47\;m/s\).


 Example 2.

A cart full of goods has a mass of \(180-kg\). To put it on the back of the delivery truck, the delivery man placed a ramp \(2.5\;m\)\(35^\circ\) above the ground. How much work is done by the man if he exerted a \(300-N\) force to push the cart and place it on the truck?How much time does it take for the man to reach the back of the truck?

Given:

\(m=180\;kg\\ d=2.5\;m\\ \theta=35^\circ\\ F=300\;N\)


Solution 1.

To solve for work at an angle, the formula \(\vec{F}\times \cos \theta\) is used.

\(W=(300\;N)(2.5\;m)\cos 35^\circ\\ \quad\,= 614.36\;J\)


Solution 2.

Compute for the velocity of the object using \(v=\sqrt \frac {2W}{m}\).

 

\(v=\sqrt \frac {2(614.36\;J)}{180\;kg}\\ \;\,=\sqrt \frac {1,228.72\;J}{180\;kg}\\ \;\,=\sqrt {6.83\;m^2/s^2}\\ \;\,=2.61\;m/s\)  


Solution 3.

From the formula of velocity, \(v=\frac dt\), derive the formula for time, \(t\).

\(v=\frac dt \implies t=\frac dv\)

 

\(t=\frac {2.5\;m}{2.61\;m/s}=0.96\;s\)

It takes \(0.96\;s\) to place the cart at the back of the truck.