A particle moves from position- r1 - 3 - - 2- -6 k- to position r2 - 14 - - 13 - - 9 k- under the action of force 4 - - - 3k- N- Calculate the work done-

Here, r_1 = 3 î + 2ĵ 6 k̂ and r_2 = 14 î + 13 ĵ 9 k̂ So, displacement S = r_2 − r_1 = (14 î + 13 ĵ 9 k̂) (3 î + 2ĵ 6 k̂) = (11 î + 11 ĵ 3 k̂) Hence, ...

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Standard XII

Physics

Work Done as Dot Product

A particle mo...

Question

A particle moves from position, ${\to r}_{1} = 3^i + 2^j - 6^k$ to position ${\to r}_{2} = 14^i + 13^j - 9^k$ under the action of force $(4^i +^j + 3^k)$ $N$ . Calculate the work done.

51 J

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100 J

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46 J

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42 J

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Solution

The correct option is C $46 J$
Here, ${\to r}_{1} = 3^i + 2^j - 6^k$ and ${\to r}_{2} = 14^i + 13^j - 9^k$
So, displacement $\to S = {\to r}_{2} - {\to r}_{1} = (14^i + 13^j - 9^k) - (3^i + 2^j - 6^k) = (11^i + 11^j - 3^k)$
Hence, Work done $W = \to F . \to S = (4^i +^j + 3^k) . (11^i + 11^j - 3^k) = 44 + 11 - 9 = 46 J$

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A particle moves from position, →r1=3^i+2^j−6^k to position →r2=14^i+13^j−9^k under the action of force (4^i+^j+3^k) N. Calculate the work done.

The correct option is C 46 JHere, →r1=3^i+2^j−6^k and →r2=14^i+13^j−9^k So, displacement →S=→r2−→r1=(14^i+13^j−9^k)−(3^i+2^j−6^k)=(11^i+11^j−3^k) Hence, Work done W=→F.→S=(4^i+^j+3^k).(11^i+11^j−3^k)=44+11−9=46 J

A particle moves from position, ${\to r}_{1} = 3^i + 2^j - 6^k$ to position ${\to r}_{2} = 14^i + 13^j - 9^k$ under the action of force $(4^i +^j + 3^k)$ $N$ . Calculate the work done.