A force F-3-4-N acts on 2kg movable object that moves from an initial position r-1-3-2-m to a final position r-1-5-4-m in 6s- The average power delivered by the force during the interval of 6s is equal to -

The correct option is A 8 watt #160; #10; #10;Given, #10; #10;Force, F⃗=(3î+4ĵ)N #10; #10;Displacement, d⃗=r⃗_2 r⃗_1=(5î+4ĵ) ( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Multiplication with Vectors

A force F⃗=...

Question

A force $\to F = (3^i + 4^j) N$ acts on $2 k g$ movable object that moves from an initial position ${\to r}_{1} = (- 3^i - 2^j) m$ to a final position ${\to r}_{1} = (5^i + 4^j) m$ in $6 s$ . The average power delivered by the force during the interval of $6 s$ is equal to :

8 w a t t

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{50}{6} w a t t

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15 w a t t

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{50}{3} w a t t

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $8 w a t t$

Given,

Force, $\to F = (3^i + 4^j) N$

Displacement, $\to d = {\to r}_{2} - {\to r}_{1} = (5^i + 4^j) - (- 3^i - 2^j) = (8^i + 6^j) m$

Work, $W = \to F . \to d = (3^i + 4^j) (8^i + 6^j) = 48 J$

Average power $P = \frac{W}{t} = \frac{48}{6} = 8 W$

Average power is $8 W$

Suggest Corrections

Similar questions

A particle moves from a point ${\to r}_{1} = 2^i + 3^j (m)$ to another point ${\to r}_{2} = 3^i + 2^j (m)$ during which a certain force $\to F = 5^i + 5^j$ acts on it. Find the work done by the force on the particle during the displacement.

Q. A force

\to F = (5^i + 4^j) N

displaces the body by

\to s = (3^i + 5^k) m

3 s .

Power generated will be

Q. A force

\to F = (2^i + 3^j) N

acts ona particle and displaces it form initial position

_{1} = (3^i + 4^j) m

to final position

_{2} = (2^i + 3^j) m

, then work done by force, is

Q. An object is displaced from position vector

{\to r}_{1} = (2^i + 3^j) m

{\to r}_{2} = (4^i + 6^j)

m under a force

\to F = (3 x^{2}^i + 2 y^j) N .

Find the work done by this force.

Q. A particle moves from a point

_{1} = (2^i + 3^j)

to another point

_{2} = (3^i + 2^j)

during which a certain force

\to F = (5^i + 5^j)

acts on it. Calculate work done by the force on the particle during this displacement.

Join BYJU'S Learning Program

A force →F=(3^i+4^j)N acts on 2kg movable object that moves from an initial position →r1=(−3^i−2^j)m to a final position →r1=(5^i+4^j)m in 6s. The average power delivered by the force during the interval of 6s is equal to :

The correct option is A 8 watt Given, Force, →F=(3^i+4^j)N Displacement, →d=→r2−→r1=(5^i+4^j)−(−3^i−2^j)=(8^i+6^j)m Work, W=→F.→d=(3^i+4^j)(8^i+6^j)=48J Average power P=Wt=486=8W Average power is 8W

A force $\to F = (3^i + 4^j) N$ acts on $2 k g$ movable object that moves from an initial position ${\to r}_{1} = (- 3^i - 2^j) m$ to a final position ${\to r}_{1} = (5^i + 4^j) m$ in $6 s$ . The average power delivered by the force during the interval of $6 s$ is equal to :