Gujarati
5.Work, Energy, Power and Collision
normal

A particle moves from position $3\hat i + 2\hat j - 6\hat k$ to $14\hat i + 13\hat j + 9\hat k$ due to a uniform force of $(4\hat i + \hat j + 3\hat k)\,N.$ If the displacement in meters then work done will be.........$J$

A$100$
B$200$
C$300$
D$250$

Solution

(a) $S = \overrightarrow {{r_2}} – \overrightarrow {{r_1}} $
$W = \overrightarrow F \,.\,\overrightarrow S $
$ = (4\hat i + \hat j + 3\hat k)\,.\,(11\hat i + 11\hat j + 15\hat k)$
$ = (4 \times 11 + 1 \times 11 + 3 \times 15)\, = 100\,J.$
Standard 11
Physics

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