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

Explanation:

Below is an attachment containing the solution.

Answer 2

Answer:

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

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

$Power,P=\frac{energy}{time} \\P=\frac{mgh+1/2mv^{2}}{t}$

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

$p=\frac{mgh}{t} \\p=\frac{1150*9.81*100sin30}{12}\\p=47*10^{3}W$

b.

for a change in velocity of 30m/s

we have the power to be

$P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*30^2}{12}\\ P=47kw +43.1kw\\P=90.1KW$

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

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