A car of weight W is on an inclined road that | vedclass

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(C) The slope of the road is given by $\sin 	heta \approx 	an 	heta = \frac{100 \, m}{1000 \, m} = \frac{1}{10}$.When moving uphill at speed $v_1 =

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$15$

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(C) The slope of the road is given by $\sin 	heta \approx 	an 	heta = \frac{100 \, m}{1000 \, m} = \frac{1}{10}$.When moving uphill at speed $v_1 = 10 \, m/s$,the force required is $F_{up} = W \sin 	heta + f = W(\frac{1}{10}) + \frac{W}{20} = \frac{3W}{20}$.The power required is $P = F_{up} \cdot v_1 = (\frac{3W}{20}) \cdot 10 = \frac{3W}{2}$.When moving downhill at speed $v$,the force required is $F_{down} = f - W \sin 	heta = \frac{W}{20} - \frac{W}{10} = -\frac{W}{20}$.The negative sign indicates that the car needs to apply a braking force or that the gravity component is greater than the friction. However,the problem states the car needs power $P/2$ to move downhill,implying the engine is working to maintain speed against the net force. The power required is $P' = |F_{net}| \cdot v = |W \sin 	heta - f| \cdot v = |\frac{W}{10} - \frac{W}{20}| \cdot v = \frac{W}{20} \cdot v$.Given $P' = \frac{P}{2} = \frac{3W}{4}$,we have $\frac{W}{20} \cdot v = \frac{3W}{4}$.Solving for $v$: $v = \frac{3 \cdot 20}{4} = 15 \, m/s$.

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