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Angelina_Jolie [31]

4 years ago

14

What is the work done by a car's braking system when it slows the 1500-kg car from an initial speed of 96 km/h down to 56 km/h i

n a distance of 55 m?

2 answers:

kondor19780726 [428]4 years ago

7 0

Answer:

-352.275KJ

Explanation:

We are given that

Mass of car=1500kg

Initial speed of car =u=96 km/h= $96\times \frac{5}{18}=26.67m/s$

1km/h= $\frac{5}{18}m/s$

Final speed of car=v=56km/h= $56\times \frac{5}{18}=15.56m/s$

Distance traveled by car=s=55m

We have to find the work done by the car's braking system.

Using third equation of motion

$v^2-u^2=2as$

$(15.56)^2-(26.67)^2=2a(55)$

$-469.18=110a$

$a=\frac{-469.18}{110}=-4.27m/s^2$

Where negative sign indicates that velocity of car decreases.

Work done by a car's barking system= $w=F\times s=ma\times s$

Work done by a car's barking system= $1500\times -4.27\times 55=352275J$

Work done by a car's barking system= $-\frac{352275}{1000}=-352.275KJ$

1KJ=1000J

Where negative sign indicates that work done in opposite direction of motion.

Gennadij [26K]4 years ago

7 0

Answer:

Work done by the car's braking system W = 3.519×10^5 J

Explanation:

we can apply newton's equation of motion to find acceleration

as

apply v^2 -u^2 = 2as

v= final velocity

u= initial velocity

also 1 kmph = 5/18 m/s

v= 56×5/18= 15.6 m/s

u= 96×5/18 = 26.66 m/s

s= 55 m

so a = [(26.66)^2 -(15.4)^2]/2×55

a = 4.26 m/s^2

so F = ma

Mass m= 1500 kg

F = 1500×4.26

F = 6397.306 N

Now work done W = F×s

W = 6397.306×55

W = 3.519×10^5 J

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