Work done by a force F- on a body is W- F-s- where s- is the displacement of body- Given that under a force F- 2-3-4k- N a body is displaced from position vector r-1- 2-3-k- m to the position vector r-2- -k- m- Find the work done by this force-

The correct option is A 8 JThe body is displaced from #160; r⃗_⃗1⃗ to r⃗_⃗2⃗. #160;Therefore, displaeement of the body is, s⃗= r⃗_⃗2⃗ r⃗_⃗1⃗= ( ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Work Done: Spring

Work done by ...

Question

Work done by a force $\to F$ on a body is $W = \to F \cdot \to s,$ where $\to s$ is the displacement of body. Given that under a force $\to F = (2^i + 3^j + 4^k) N$ a body is displaced from position vector $_{1} = (2^i + 3^j +^k)$ m to the position vector $_{2} = (^i +^j +^k) m .$ Find the work done by this force.

-8 J

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

8 J

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

-16 J

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

16 J

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

Open in App

Solution

The correct option is A -8 J
The body is displaced from $_{1} t o_{2} .$
Therefore, displaeement of the body is,
$\to s =_{2} -_{1} = (^i +^j +^k) - (2^i + 3^j +^k) = (-^i - 2^j) m$
Now, work done by the force is:
$W = \to F \cdot \to s$
$= (2^i + 3^j + 4^k) \cdot (-^i - 2^j)$
$= (2) (- 1) + (3) (- 2) = - 8 J$

Suggest Corrections

Similar questions

Q. An object is displaced from position vector

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

_{2} = (4^i + 6^j) m

under the action of a force

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

. Find the work done by this force.

An object has a displacement from position vector $_{1} = (2^i + 3^j) m$
to $_{2} = (4^i + 6^j) m$ under a force $\to F = (3 x^{2}^i + 2 y^j) N,$ then work done by the force is:

Q. If

\to F = 2^i + 3^j + 4^k

acts on a body and displaces it by

\to S = 3^i + 2^j + 5^k

, then the work done by the force is:

A

force

\to F = 2^i + 4^j +^k

acts on a body and produces a displacement of

\to S = 3^i + 2^j + 5^k

. Work done is:

Q. A body is displaced from

{\to r}_{A} = (2 m, 4 m, - 6 m)

_{B} = (6^i - 4^j + 2^k)

m under a constant force

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

Find the work done.

Join BYJU'S Learning Program

Work done by a force →F on a body is W=→F⋅→s, where →s is the displacement of body. Given that under a force →F=(2^i+3^j+4^k)N a body is displaced from position vector →r1=(2^i+3^j+^k) m to the position vector →r2=(^i+^j+^k)m. Find the work done by this force.

The correct option is A -8 JThe body is displaced from →r1to→r2. Therefore, displaeement of the body is,→s=→r2−→r1=(^i+^j+^k)−(2^i+3^j+^k)=(−^i−2^j)mNow, work done by the force is: W=→F⋅→s =(2^i+3^j+4^k)⋅(−^i−2^j) =(2)(−1)+(3)(−2)=−8J