A toy of mass 20 g at rest acquires a velocity (3i^−2j^) ms−1 in 2 s. Then, the power of the toy is

Power=Work done by forceTime=change in KEtime=12mv2t As object is initially at rest, change in KE = 12mv2 ⇒ Power=12×201000×(3i^−2j^)22 =20×(9+4)4000=13200=6.5100=0.065W

A toy of mass 20 g at rest acquires a velocity (3i^−2j^) ms−1 in 2 s. Then, the power of the toy is

Power=Work done by forceTime=change in KEtime=12mv2t As object is initially at rest, change in KE = 12mv2 ⇒ Power=12×201000×(3i^−2j^)22 =20×(9+4)4000=13200=6.5100=0.065W

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A toy of mass 20 g at rest acquires a velocity $(3 \hat{i} - 2 \hat{j}) m s^{- 1}$ in 2 s. Then, the power of the toy is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

0.975 W

0.325 W

1.3 W

0.065 W

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$P o w e r = \frac{W o r k d o n e b y f o r c e}{T i m e} = \frac{c h a n g e i n K E}{t i m e} = \frac{\frac{1}{2} m v^{2}}{t}$
As object is initially at rest, change in KE = $\frac{1}{2} m v^{2}$
$\Rightarrow P o w e r = \frac{\frac{1}{2} \times \frac{20}{1000} \times {(3 \hat{i} - 2 \hat{j})}^{2}}{2}$
$= \frac{20 \times (9 + 4)}{4000} = \frac{13}{200} = \frac{6.5}{100} = 0.065 W$