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25 November, 06:20

A 1150 kg car climbs a 100 m road that has a slope of 30o in 12 s. Calculate the power required for: Question 3.1: Constant velocity. Question 3.2: From rest (zero velocity) to 30 m/s. Question 3.3: From 35 m/s to 5 m/s.

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  1. 25 November, 09:59
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    47 KW

    90.1 KW

    -10.5 KW

    Explanation:

    Knowns and requirements

    m=1150 kg-> mass of the car

    L=100 m-> road length

    It's required to determine the power needed to climb the uphill road (30° from horizontal) in 12 second

    a) at constant speed

    b) from rest to a final velocity of 30 m/s

    c) from 35 m/s to a final velocity of 5 m/s

    a) The power required to climb the uphill road can simply expressed as the rate of change of the kinetic energy and potential energy

    P=m*g*Δh/Δt+0.5m (v_2^2-v_1^2) / Δt (1)

    where Δh is the vertical rise of the road (Δh = L sin (30))

    Since the car moving at constant speed, there will be no change in kinetic energy and Eq. (1) reduced to

    P=m*g*Δh/Δt

    =1150*9.8*100*sin (30) / 12s

    =47 KW

    b) Bulging our values (v_1 = 0.0) and (v_2 = 30 m/s) into Eq. (1), The required power will be

    P=m*g*Δh/Δt+0.5m (v_2^2-v_1^2) / Δt

    =90.1 KW

    c) Bulging our values (v_1 = 35) and (v_2 = 5 mis) into Eq. (1), The required power will be

    P=m*g*Δh/Δt+0.5m (v_2^2-v_1^2) / Δt

    =-10.5 KW

    Note : The negative sign means that the car is losing energy. In other words the driver must use the brake to reduce the velocity from 35 mis to 5 mis during climbing the uphill road.
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