Question #116760

a car has a drag coefficient Cd=0.30, a frontal area of A = 1.9 meter square, a mass 1.2 tonnes and a coefficient of rolling resistance, Cr,= 0.012. it is travelling up a hill with a slope of 1 in 20 at 110 kph.

at what rate it is doing work against gravity(i.e. at what rate is it increasing its gravitational potential energy)?

Pg=_______kW.

(a 1:20 grade means that it rises 1m for every 20 m travelled along the road: sin(theta) = 1/20.)

Expert's answer

Pg=ΔmgHΔt=mgΔHΔtP_g=\frac{\Delta mgH }{\Delta t}=mg\frac{\Delta H}{\Delta t} as ΔH=v(sinθ)Δt\Delta H=v(\sin \theta) \Delta t then Pg=mgVsinθ1.21039.81103.60.05P_g=mgV\sin\theta\approx1.2\sdot10^3\sdot9.8\sdot110\sdot3.6\sdot0.05\approx

232.8103=233kW\approx232.8\sdot10^3=233kW

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