Two constant forces {F_1} = 2hat i - 3hat j + | vedclass

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Question

(A) The net force acting on the body is $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = (2\hat i - 3\hat j + 3\hat k) + (\hat i + \hat j - 2\hat k) = 3\hat i

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The net force acting on the body is $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = (2\hat i - 3\hat j + 3\hat k) + (\hat i + \hat j - 2\hat k) = 3\hat i - 2\hat j + \hat k$ $(N)$.The displacement of the body is $\vec{d} = \vec{r}_2 - \vec{r}_1 = (7\hat i + 10\hat j + 5\hat k) - (\hat i + 2\hat j - 2\hat k) = 6\hat i + 8\hat j + 7\hat k$ $(m)$.The work done is given by the dot product of the net force and the displacement: $W = \vec{F}_{net} \cdot \vec{d}$.$W = (3\hat i - 2\hat j + \hat k) \cdot (6\hat i + 8\hat j + 7\hat k) = (3 	imes 6) + (-2 	imes 8) + (1 	imes 7) = 18 - 16 + 7 = 9$ $J$.

Two constant forces ${F_1} = 2\hat i - 3\hat j + 3\hat k$ $(N)$ and ${F_2} = \hat i + \hat j - 2\hat k$ $(N)$ act on a body and displace it from the position ${r_1} = \hat i + 2\hat j - 2\hat k$ $(m)$ to the position ${r_2} = 7\hat i + 10\hat j + 5\hat k$ $(m)$. What is the work done in $J$?

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