A time dependent force F=10t is applied on 10kg .Find work done by F in 2 secs
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F=ma =10a
F=10 *2 =20
so 20= 10 a
a=2 m/s^2
work =force*distance
s=ut+1/2 at^2
s= 0+1/2 2[2]^2
= 4
so, w= 20 * 4 =80 ans
F=10 *2 =20
so 20= 10 a
a=2 m/s^2
work =force*distance
s=ut+1/2 at^2
s= 0+1/2 2[2]^2
= 4
so, w= 20 * 4 =80 ans
Anonymous:
is it wrong ?
Yes answer is 20J
Also how F=10*2
F=10 t and t = 2 sec
is my ans ryt / wrng
if it helped , plz mark it brainliest ., ^_^
wrng
oh then srry :(
np
F= ma = 10 * dv/dt......t= dv/dt ( mass is 10kg..both get cancelled).... Integration( t dt) = integration( dv)....... t^2/2 = v...... v= 4/2 ( t=2s).....v= 2m/s.....v= u + at..... a = 2/2....a= 1m/s^2...... S = ut + 1/2 at^2........ S= 1/2 * 1 * 4 ( u=0m/s)..... S = 2m.........WORK = F * S...... WORK = 10 * 2.....Therefore work = 20J.....Cheers! Hope it helped
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