A body of mass 1kg begins to move under the action of a time dependent force F→=2ti^+3t2j^N , where i^ and j^ are unit vectors along x and y - axis. What power will be developed by the force at the time t ?

Given force F→=2ti^+3t2j^According to Newton’s second law of motion,mdv→dt=2ti^+3t2j^(m=1kg)⇒∫0v→ dv→=∫0t 2ti^+3t2j^dt⇒v→=t2i+t3j^Power P=F→⋅v→2ti^+3t2j^⋅t2i^+t3j^(2t3 + 3t5) W

A body of mass 1kg begins to move under the action of a time dependent force F→=2ti^+3t2j^N , where i^ and j^ are unit vectors along x and y - axis. What power will be developed by the force at the time t ?

Given force F→=2ti^+3t2j^According to Newton’s second law of motion,mdv→dt=2ti^+3t2j^(m=1kg)⇒∫0v→ dv→=∫0t 2ti^+3t2j^dt⇒v→=t2i+t3j^Power P=F→⋅v→2ti^+3t2j^⋅t2i^+t3j^(2t3 + 3t5) W

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 body of mass 1kg begins to move under the action of a time dependent force $\vec{F} = (2 t \hat{i} + 3 t^{2} \hat{j}) N$ , where $\hat{i} and \hat{j}$ are unit vectors along x and y - axis. What power will be developed by the force at the time t ?

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

(2t³ + 3t⁵) W

(2t² - 3t³) W

(2t³ + 3t⁴) W

(2t² + 4t⁴) W

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given force $\vec{F} = 2 t \hat{i} + 3 t^{2} \hat{j}$
According to Newton’s second law of motion,
$\begin{array}{l} m \frac{d \vec{v}}{dt} = 2 t \hat{i} + 3 t^{2} \hat{j} (m = 1 kg) \\ \Rightarrow \int_{0}^{\vec{v}} d \vec{v} = \int_{0}^{t} (2 t \hat{i} + 3 t^{2} \hat{j}) dt \\ \Rightarrow \vec{v} = t^{2} i + t^{3} \hat{j} \end{array}$
Power $P = \vec{F} \cdot \vec{v} (2 t \hat{i} + 3 t^{2} \hat{j}) \cdot (t^{2} \hat{i} + t^{3} \hat{j})$
(2t³ + 3t⁵) W