Given, constant applied force, F=5N, Mass of particle, m=500g=‌
500
1000
=0.5kg Displacement, s=5m Initial speed, u=0ms−1 Final speed =v As we know that, Power(P)=‌
‌ Work ‌(W)
Time(t)
. . . (i) According to work-energy theorem, Work done, W=‌
1
2
m(v2−u2) . . . (ii) and F=ma where, a is acceleration. ∴‌‌a‌‌=‌
F
m
‌‌=‌
5
0.5
=‌
50
5
=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=‌
1
2
×0.5(102−02) ‌‌=‌
1
4
(100)=25J . . . (iii) According to first equation of motion, v=u+at ⇒‌‌t=‌
v−u
a
⇒‌‌t=‌
10−0
10
=1s . . . (iv) Now, from Eqs. (i), (iii) and (iv), we get P‌‌=‌