A truck of mass 1800kg is moving with a speed of 54Km/h when brakes are applied its stops with uniform negative acceleration at a distance of 200m. calculate the force applied by the brakes of the truck and work done before stopping.
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Answered by
119
m=1800 kg
u = 54 kmph = 15 m/s
v=0
s=200 m
a=v²-u²/2s
a=0-225/400
a=-0.5625m/s²
f=ma
f=1800*-0.5625
f=1012.5 N
w=fs
w=1012.5*200
w=202500 J
u = 54 kmph = 15 m/s
v=0
s=200 m
a=v²-u²/2s
a=0-225/400
a=-0.5625m/s²
f=ma
f=1800*-0.5625
f=1012.5 N
w=fs
w=1012.5*200
w=202500 J
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30
The initial velocity of the truck is,
u=54 km/hru=54×518 m/su=15 m/s
v2−u2=2as
(0)2−(15 m/s)2=2a×200 m−(15 m/s)2=a×400 ma=−225 m2/s2400 ma=−0.56 m/s2
F=ma
F=(1800 kg)×(−0.56 m/s2)F=−1012.5 N
F=1012.5 N
W=Fs
W=1012.5 N×200 mW=202500 N.m
u=54 km/hru=54×518 m/su=15 m/s
v2−u2=2as
(0)2−(15 m/s)2=2a×200 m−(15 m/s)2=a×400 ma=−225 m2/s2400 ma=−0.56 m/s2
F=ma
F=(1800 kg)×(−0.56 m/s2)F=−1012.5 N
F=1012.5 N
W=Fs
W=1012.5 N×200 mW=202500 N.m
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