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Asked by Yashraj Last Modified
Answer 
mass=1000kg
u=0km/hr=0m/s
v=54km/hr=15m/s
work= change in kinetic energy
= 1/2xmxv^2 - 1/2xmxu^2
= 1/2x1000x15^2 - 0
=112500 J
power = work/time
= 112500/5 = 22500 J/s=22.5 kw
from first eqution of motion
v=u+at
u=0
therfore
v=at
15=5a
a=3m/s^2
from second equation of motion
s=ut+/2at^2
0+1/2*3*5*%
s=37.5 m
now v avg=dsance/time=37.5/5=7.5 m/s
gaain in KE=KE fina=KE initial
=1/2mv^2-1/2mv2^2
=1/2*1000*15^=112.5kj
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m = 1000kg
u(initial velocity) = 0 ; as the body starts from rest .
v( final velocity ) = 54 km/h = 15 m/s
t(time) = 5 sec
using equation of motion ,
v = u + at , or
15 = 0 + a*5, or
a = (15-0) / 5
so , a = 3 m /s^2
now , we know kinetic energy = 1/2 M V^2
so K.E = 1/2 * 1000 *15 *15 = 112500 Joules .
read lessRelated Questions
The moment of inertia of a body about a given axis is 1.2kgm^2 . Initially the body is at rest. In order to produce a rotational kinetic energy of 1500J, an angular acceleration of 25rad/second ^2 must be applied about that axis for a duration of
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