Question

A block of mass 5 kg is thrown upwards with a velocity of 10 m/s. It momentarily comes to rest at a height of 4 m. What is the work done by air friction durring this interval?

Answer 1

$$\begin{gathered} Kineticenergy=workdonebygravity+workdonebyfriction \\ 0.5(mv{}^2)=mgh+workdonebyfriction \\ 0.5*5*{10}^2=5*10*4+workdonebyfriction \\ 250-200=workdonebyfriction \\ 50Joule=workdonebyfriction \end{gathered}$$