Yahoo Answers: Answers and Comments for Physics question work power energy please help... please show the working? [Physics]
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From vidit g
enUS
Mon, 22 Mar 2010 10:15:41 +0000
3
Yahoo Answers: Answers and Comments for Physics question work power energy please help... please show the working? [Physics]
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https://answers.yahoo.com/question/index?qid=20100322101541AAc92pm
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From Kat: Q1
a=(k^2)r(t^2)
using F=ma
F/m =(k^2)r(t...
https://answers.yahoo.com/question/index?qid=20100322101541AAc92pm
https://answers.yahoo.com/question/index?qid=20100322101541AAc92pm
Sun, 28 Mar 2010 20:10:40 +0000
Q1
a=(k^2)r(t^2)
using F=ma
F/m =(k^2)r(t^2)
F=m(k^2)r(t^2)
E= F x r (Energy=force x distance)
F=E/r
So E/r= m(k^2)r(t^2)
E=m(k^2)(r^2)(t^2)
Power =Energy/time =E/t
Therefore P=m(k^2)(r^2)t
Q2
distance travelled,s = ut +1/2a(t^2) where u=initial velocity, a =acceleration, t=time
Assuming u= 0
s=1/2a(t^2)
F=ma
F = power/velocity
P/v =ma , but P and m are constant so;
1/v is proportional to a
v =s/t
1/v =t/s
t/s proportional to a
Substitute t/s for a in s=1/2a(t^2)
s is proportional to (t/s) *(t^2)
s is proportional to (t^3)/s
so s^2 is proportional to t^3
and s is proportional to t^3/2