Answer:
Work done by external force is given as
![Work_(external) = mgLsin\theta + \mu mgLcos(\theta) + (1)/(2)mv_2^2 - (1)/(2)mv_1^2](https://img.qammunity.org/2020/formulas/physics/high-school/1bphggzhdgdmephagcl22ukb2uomoxjmlt.png)
Step-by-step 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
![Work_(external) + Work_(gravity) + Work_(friction) = (1)/(2)mv_2^2 - (1)/(2)mv_1^2](https://img.qammunity.org/2020/formulas/physics/high-school/mrarruibr08rmgyw7nipbh0hrn2f2ku4s8.png)
now we have
![W_(gravity) = -mg(Lsin\theta)](https://img.qammunity.org/2020/formulas/physics/high-school/r9v7sfrox110vac0hdx77qewhglzd2wipn.png)
![W_(friction) = -\mu mgcos(\theta) L](https://img.qammunity.org/2020/formulas/physics/high-school/tteo48vzlyuqmcoekytjsd21h2errf8tvy.png)
so from above equation
![Work_(external) - mgLsin\theta - \mu mgLcos(\theta) = (1)/(2)mv_2^2 - (1)/(2)mv_1^2](https://img.qammunity.org/2020/formulas/physics/high-school/mmzv0lg77pjf8a36blnilh90ti6sxg2n1j.png)
so from above equation work done by external force is given as
![Work_(external) = mgLsin\theta + \mu mgLcos(\theta) + (1)/(2)mv_2^2 - (1)/(2)mv_1^2](https://img.qammunity.org/2020/formulas/physics/high-school/1bphggzhdgdmephagcl22ukb2uomoxjmlt.png)