A car of mass m accelerates with acceleration a up an inclined plane of angle th

Question

A car of mass m accelerates with acceleration a up an inclined plane of angle theta as in Fig. 6.41. The drag force f_D consists of rolling friction alpha (N) and air drag beta v^2 (N) i.e. f_D = alpha + beta v^2, where alpha and beta are constants and v is the speed of the car. (a) Find the force F that propels the car. Show that P = mva + mvg sin theta + alpha v + beta v^3 is the power delivered to the wheels by the engine, where mva is the power delivered to accelerate the car, mvg sin theta is the power to overcome gravity, alpha v is the power to overcome rolling friction, and beta v^2 is the power to overcome air drag. (c)Calculate the various components of P and hence the total P if we take m = 1,000 kg, a = 2 m/s^2 v = 20 m/s alpha = 200N, beta = 0.5 kg/m, and theta = 15 degree.

Explanation / Answer

a) The force up the incline is F and forces down the incline on the car are mgsin and fD = + v2. The car goes up the incline with an acceleration a. So, the equation of motion of car up the incline of angle is given as;

F - mgsin - fD = ma

or F - mgsin - - v2 = ma, where v is the speed of car

or F = mgsin + + v2 + ma ------------ (1)

This F is the force which propels the car.

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b) Let at an instant the speed of the car be v.

Then power delivered is P = F.v = Fvcos00 = Fv (F and v are both up the incline, so angle between them is zero)

or P = Fv = (mgsin + + v2 + ma)v

or P = mva + mvgsin +  v + v3

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c) the various compnents of P are

mva = (1000kg)(20m/s)(2m/s2) = 40,000W

mvgsin = (1000kg)(20m/s)(9.8m/s2)(sin150) = 50728.53W

v = (200N)(20m/s) = 4000W

v3 = (0.5kg/m)(20m/s)3 = 4000W

Hence the total P is

P = mva + mvgsin +  v + v3 = 40,000W + 50728.53W + 4000W + 4000W

P = 98728.53W

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This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige.....

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