A car is moving on an inclined road with friction acting upon it. When it is moving up the road at a speed v the engine is working at power 3P and when it is moving down the road at v the engine is working at a power P. Find the value of P.

Incline is at θ where sin θ = 1/20 and mass of the car is 800kg and v is 12.5 m/s

Up the road: Power = Fv                              F = R + (800g)/20

                                             Power = (R + 40g)*25/2 = 3P …. P = (R + 40g)*25/6

Down the road: Power = Fv          F = R – (800g)/20

                                             Power = (R - 40g)*25/2 = P

Equate the equations:

(R + 40g)*25/6 = (R – 40g)*25/2

25R + 1000g = 75R – 3000g so R = 80g

Therefore P = (80g – 40g)*25/2 = 500g = 4900W

Answered by Jonathan M. Maths tutor

2845 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What are the parameters of the Poisson distribution?


Express x^2-7x+2 in the form (x-p)^2+q where p and q are rational. Hence or otherwise find the minimum value of x^2-7x+2


Use the substition u = cos(x) to find the indefinite integral of -12sin(x)cos^3(x) dx


Find the set of values of k for which x^2 + 2x+11 = k(2x-1)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences