Question

A car of mass 1000 kg moving with a velocity of 72 km/h comes to a half in 10 sec; find the retardation and force applied by the breaks of the car?

Answer 1

Answer:

Given mass of car =1000kg

u=speed=90km/hr=90×

18

5

m/sec

u=25m/sec

it have to stop in 5sec

then final velocity =0

⇒v=u+at

o=25+a×5

=a=−5m/sec

2

then the force applied by break is ma

=1000X−(5m/sec

2

)

=−5000M

force =5000N by break

distance travelled to stop the car after break is applied (s=distance)

v

2

=u

2

+2as

0=(25)

2

−2×5×s

=s=

10

(25)

2

=62.5m

work done by break =F×s

w=5000×62.5=312.5KJ

Step-by-step explanation:

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Answer 2

M=1000Kg

U (initial) =72km/h ={(72*1000) /3600}m/s=20m/s

V (final) =72/2=36km/h={(36*1000) /3600}m/s=10m/s

we need to find retardation and force

so V=U+at

10= 20 + a*10

10-20=a*10

-10/10=a

a= -1m/s^2

so Retardation=1m/s^2

and force is

F=ma

F=(1000kg)*(-1m/s^2)

F=-1000 Newton

- sign indicates force is in opposite dir. of the motion of the car