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Question Number 177922 by Lekhraj last updated on 11/Oct/22

Answered by mr W last updated on 11/Oct/22

$$\left(i\right)$$$$W=mgs\sin \theta +\frac{m(v_2^2-v_1^2)}{2}$$$$=700\times 9.81\times 400\times \sin 5°+\frac{700({15}^2-{20}^2)}{2}$$$$\approx 178149J\approx 178kJ$$$$\left(ii\right)$$$$P_2=(mg\sin \theta +R)v_2$$$$=(700\times 9.81\times \sin 5°+200)\times 15$$$$\approx 11977W\approx 12.0kW$$$$\left(iii\right)$$$$P_1=({ma}_1+mg\sin \theta +R)v_1=P_2$$$$\Rightarrow a_1=\frac{P_2}{{mv}_1}-g\sin \theta -\frac{R}{m}$$$$=\frac{11977}{700\times 20}-9.81\times \sin 5-\frac{200}{700}$$$$\approx -0.285m/s^2$$

Commented by Lekhraj last updated on 11/Oct/22

$$thankyou$$

Commented by Tawa11 last updated on 11/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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