A force F -3 - cj -2 k- N acting on a particle causes a displacement of S -4 -2 -3 k- m- If the work done is 6 J- then the value of- c - isA- 1B- 6C- 12 D -0

Given: F = 3 î + c ĵ + 2 k̂, and S = 4 î + 2ĵ 3k̂ Work done is given by, W = F·S Now according to question, W = F·S = 6 ......(1) Using (1), we get, ( ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

A force F =3 ...

Question

A force $F = (3^i + c^j + 2^k) N$ acting on a particle causes a displacement of $\to S = (- 4^i + 2^j - 3^k) m$ . If the work done is $6 J$ , then the value of $^{'} c^{'}$ is

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Solution

The correct option is D $12$
Given
$F = 3^i + c^j + 2^k$ and $S = - 4^i + 2^j - 3^k$
Work done
$W = \to F . \to S$
Now according to question, $W = \to F . \to S = 6$ ...... $(1)$

Using $(1)$ , we get $(3^i + c^j + 2^k) . (- 4^i + 2^j - 3^k) = 6$
$\Rightarrow 3 \times (- 4) + 2 c + 2 \times (- 3) = 6 \Rightarrow 2 c = 24 \Rightarrow c = 12$

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A force F=(3^i+c^j+2^k) N acting on a particle causes a displacement of →S=(−4^i+2^j−3^k) m. If the work done is 6 J, then the value of ′c′ is

The correct option is D 12Given F=3^i+c^j+2^k and S=−4^i+2^j−3^k Work done W=→F.→S Now according to question, W=→F.→S=6 ......(1) Using (1), we get (3^i+c^j+2^k).(−4^i+2^j−3^k)=6 ⇒3×(−4)+2c+2×(−3)=6⇒2c=24⇒c=12

A force $F = (3^i + c^j + 2^k) N$ acting on a particle causes a displacement of $\to S = (- 4^i + 2^j - 3^k) m$ . If the work done is $6 J$ , then the value of $^{'} c^{'}$ is