Amount of Work Done = Amount of Energy Converted
Example:
A trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms-1 and it stops at B after 4 seconds. What is the work done to overcome the friction?
Answer:
In this case, kinetic energy is converted into heat energy due to the friction. The work done to overcome the friction is equal to the amount of kinetic energy converted into heat energy, hence
\[\begin{array}{l}{\rm{Work Done}} \\{\rm{ = Kinetic Energy Loss}} \\= \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 \\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}(5)(4)^2 - \frac{1}{2}(5)(0)^2 \\= 40J \\\end{array}\]
A trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms-1 and it stops at B after 4 seconds. What is the work done to overcome the friction?
Answer:
In this case, kinetic energy is converted into heat energy due to the friction. The work done to overcome the friction is equal to the amount of kinetic energy converted into heat energy, hence
\[\begin{array}{l}{\rm{Work Done}} \\{\rm{ = Kinetic Energy Loss}} \\= \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 \\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}(5)(4)^2 - \frac{1}{2}(5)(0)^2 \\= 40J \\\end{array}\]
