Question

Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 308 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance.

Answer 1

the question is incomplete, below is the complete question

"Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 30 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance."

a.
`47KW`

b. 90.1KW

c. -10.5KW

Step-by-step explanation:

Data given

mass m=1150kg,

length,l=100m

angle, =30

time=12s

Note that the power is define as the rate change of the energy, note we consider the potiential and the

Hence

`Power,P=(energy)/(time) \\P=(mgh+1/2mv^(2))/(t)`

to determine the height, we draw the diagram of the hill as shown in the attached diagram

a. at constant speed, the kinetic energy is zero.Hence the power is calculated as

`p=(mgh)/(t) \\p=(1150*9.81*100sin30)/(12)\\p=47*10^(3)W\\`

b.

for a change in velocity of 30m/s

we have the power to be

`P=(mgh)/(t) +(1/2mv^2)/(t)\\ P=47Kw+( 0.5*1150*30^2)/(12)\\ P=47kw +43.1kw\\P=90.1KW`

c. when i decelerate, we have the power to be

`P=(mgh)/(t) +(1/2mv^2)/(t)\\ P=47Kw+( 0.5*1150*\\((5^2)-(35^2)))/(12)\\ P=47kw -57.5kw\\P=-10.5KW`

Answer 2

Step-by-step explanation:

Below is an attachment containing the solution.