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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Question:

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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Answer:47 KW90.1 KW-10.5 KWExplanation:Knowns and requirements m=1150 kg—> mass of the carL=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 KWb) 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 KWc) 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 KWNote : 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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