Answer:
Work done by external force is given as
[tex]Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
Explanation:
As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car
so we will have
[tex]Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
now we have
[tex]W_{gravity} = -mg(Lsin\theta)[/tex]
[tex]W_{friction} = -\mu mgcos(\theta) L[/tex]
so from above equation
[tex]Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
so from above equation work done by external force is given as
[tex]Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
