Answer:
[tex]-mgLsin\theta[/tex]
Explanation:
We are given that
Applied force=F
Angle=[tex]\theta[/tex]
Distance=L
We have to find the work done by the force due to gravity.
[tex]g=9.8 m/s^2[/tex]
Work done=[tex]mgsin\theta Lcos180^{\circ}+mgcos\theta Lcos90[/tex]
[tex]W_g=-mgsin\theta L+mgLcos\theta\times 0[/tex]
Where [tex]cos180^{\circ}=-1[/tex]
[tex]cos90^{\circ}=0[/tex]
[tex]W_g=-mgLsin\theta[/tex]
Hence, the work done on the block by the force of gravity as the block moves a distance L up the incline=[tex]-mgLsin\theta[/tex]
