A car of weight w is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force on the car. While moving uphill on the road at a speed of , the car needs power p. If it needs power while moving downhill at speed v then value of v is:
Answers
Your question seems wrong. a correct question is ——-> A car of weight W is on an inclined road that rises by 100m over a distance of 1km and applies a constant frictional force W/20 on the car. While moving uphill on the road at a speed of 10ms−1 the car needs power P. If it needs power P/2 while moving downhill at speed v then value of v is ....
solution : sinθ = 100m/1000m = 1/10
weight of car = W
frictional force = W/20
we know, power = net force × velocity
while moving uphill , net force = weight of body along plane + frictional force = wsinθ + w/20
so, power = (wsinθ + w/20) × 10 [ velocity of car = 10m/s ]
Power = (w/10 + w/20) × 10 = 3w/2
similarly, while moving downhill , net force = weight of body - frictional force
net force = wsinθ - w/20 = w/20
So, power = w/20 × v
P/2 = w/20 × v
3w/4 = w/20 × v
v = 15 m/s
Hence, speed of car downhill = 15m/s