Given, constant applied force,
F=5N,
Mass of particle,
m=500g==0.5kg Displacement,
s=5m Initial speed,
u=0ms−1 Final speed
=v As we know that,
Power(P)=. . . (i)
According to work-energy theorem,
Work done,
W=m(v2−u2) . . . (ii)
and
F=ma where,
a is acceleration.
∴a= ===10ms−2 and by using third equation of motion
⇒v2−u2=2as ⇒v2=2as v=√2as ∴v=√2×10×5=√100 =10ms−1 Substituting the values in Eq. (ii), we get
W=×0.5(102−02) =(100)=25J . . . (iii)
According to first equation of motion,
v=u+at ⇒t= ⇒t==1s . . . (iv)
Now, from Eqs. (i), (iii) and (iv), we get
P= ==25W
