an object (vechical) of mass 50kg moving with the speed of 20m/s break in applied on it and it stop in 5 sec. calculate work done by break?
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v=d/t
d=20*5=100m
a=20/5=4m/s^2
w=f*d
w=50*4*100=20000N/m
d=20*5=100m
a=20/5=4m/s^2
w=f*d
w=50*4*100=20000N/m
Answered by
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mass=50kg
speed=20m/s
time=5sec
distace=speed×time
=20×5
=100m
v^2-u^2=2as
o-20 =2×a×100
-20 =200a
a = -20/200
a =-1/10m/s^2
workdone=m×a×d
=50×-1/10×100
= -500j
speed=20m/s
time=5sec
distace=speed×time
=20×5
=100m
v^2-u^2=2as
o-20 =2×a×100
-20 =200a
a = -20/200
a =-1/10m/s^2
workdone=m×a×d
=50×-1/10×100
= -500j
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