A car of mass 1400kg accelerates up a road with a slope of 7.181*, increasing it

Question

A car of mass 1400kg accelerates up a road with a slope of 7.181*, increasing its F speed from 4 to 16m.s^-1, while travelling up the road a distance of 80m, against a h frictional resistance of 1.0kN. Determine: a) The work done during the period of acceleration b) The tractive effort between the driving wheels and the road surface (Total Force required) c) The average power developed For the merit Use both the work energy principle and linear motion to compare the calculations and solutions to this problem.

Explanation / Answer

here,

theta = 7.181 degree

m = 1400 kg

frictional force , ff = 1000 N

d = 80 m

let the accelration be a

using third equation of motion

v^2 - u^2 = 2 * a * d

16^2 - 4^2 = 2 * a * 80

a = 1.5 m/s^2

a)

the work done during the period of accelration , W = total energy gained

W = m * g * d * sin(theta) + 0.5 * m * ( v^2 - u^2) - ff * d

W = 1400 * ( 9.8 * 80 * sin(7.181) + 0.5 * ( 16^2 - 4^2)) - 1000 * 80

W = 2.25 * 10^5 J

the work done during accelration is 2.25 * 10^5 J

b)

the total force required , F = m * a + ff

F = 1400 * 1.5 + 1000

F = 3100 N

c)

let the time taken be t

v = u + a * t

16 = 4 + 1.5 * t

t = 8 s

the average power developed , Pavg = W/t

Pavg = 2.2 * 10^5 /8 = 2.75 * 10^4 W

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