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.

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.  

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

it was given as a lesson and guide for prayer.