Question

Two forces F = 3î – 2j+ 4k N and F = î– 3j+k
N act on a particle to cause a displacement of
î+ 3j+6k) m. Determine the total work done.​

Answer 1

$\underline{\underline{\sf \qquad Answer :\qquad}}$

As we know that, work done by the constant force (F) on particle which undergoes displacement (s) is given by :

$\bullet \:\: \underline{ \boxed{ \textsf{\textbf{Work done = \overrightarrow{\text F} . \overrightarrow{\text s} = Fs cos \theta }}}}$

Now,

$\displaystyle{ \sf:\implies s = \hat{i} + 3 \hat{j} + 6\hat{k}} \qquad\bigg\lgroup \textsf{\textbf{Given}} \bigg\rgroup$

$\displaystyle{ \sf:\implies F_{1} = 3 \hat{i} - 2 \hat{j} + 4\hat{k}}$

$\displaystyle{ \sf:\implies F_{2} = \hat{i} - 3 \hat{j} + \hat{k}}$

$\dag \: \frak{\underline{Now, let's \: calculate \: the \: total \: Force :}}$

$\displaystyle{ \sf:\implies F = F_1 + F_2}$

$\displaystyle{ \sf:\implies F =3 \hat{i} - 2 \hat{j} + 4\hat{k} +\hat{i} - 3 \hat{j} + \hat{k}}$

$\displaystyle{ \sf:\implies F = 4 \hat{i} - 5 \hat{j} + 5\hat{k}}$

$\dag \: \frak{\underline{Now, substitue \: the \: given \: values \:in above \: formula \: of \: work \: done :}}$

$\dashrightarrow\:\: \sf Work \: done = Fs \cos(\theta)$

$\dashrightarrow\:\: \sf Work \: done = (4 \hat{i} - 5 \hat{j} + 5\hat{k}).(\hat{i} + 3 \hat{j} + 6\hat{k}) \cos( {0}^{ \circ} )$

$\dashrightarrow\:\: \sf Work \: done = 4 - 15 + 30$

$\dashrightarrow\:\: \sf Work \: done = - 11 + 30$

$\dashrightarrow\:\: \underline{ \boxed{ \sf Work \: done = 19 \: J}}$

Answer 2

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♧Question♧

• Two forces F = 3î – 2j+ 4k N and F = î– 3j+k N act on a particle to cause a displacement of î+ 3j+6 km. Determine the total work done.

♧Answer♧

Given data:-

• ➠F1 = 3î – 2j+ 4k N
• ➠F2 = î– 3j+k
• ➠Displacement = î+ 3j+6 km

To find:-

• ➠Work done = ?

Solution:-

☆Force and displacement are given by:

$\:\: \red{ \boxed{ \textsf{\textbf{Work done = \overrightarrow{\text F} . \overrightarrow{\text s} = Fs cos \theta }}}}$

Now,

$\displaystyle{ \sf:\dashrightarrow s = \hat{i} + 3 \hat{j} + 6\hat{k}} \\ \\ \\{ \sf:\dashrightarrow F_{1} = 3 \hat{i} - 2 \hat{j} + 4\hat{k}} \\ \\ \\ { \sf:\dashrightarrow F_{2} = \hat{i} - 3 \hat{j} + \hat{k}}$

☆First we have to find the value of F.

$\displaystyle{ \sf:\dashrightarrow{ F = F_1 + F_2}}\\ \\ \\ { \sf:\dashrightarrow F =3 \hat{i} - 2 \hat{j} + 4\hat{k} +\hat{i} - 3 \hat{j} + \hat{k}} \\ \\ \\ { \sf:\dashrightarrow F = 4 \hat{i} - 5 \hat{j} + 5\hat{k}}$

☆Now put the values in the formula,

$\dashrightarrow\:\: \frak{Work \: done = Fs \cos(\theta)} \\ \\ \\ \dashrightarrow\:\: \frak {Work \: done = (4 \hat{i} - 5 \hat{j} + 5\hat{k}).(\hat{i} + 3 \hat{j} + 6\hat{k}) \cos( {0}^{ \circ} )} \\ \\ \\ \dashrightarrow\:\: \frak{ Work \: done = 4 - 15 + 30} \\ \\ \\ \dashrightarrow\:\: \frak {Work \: done = - 11 + 30} \\ \\ \\ \dashrightarrow\:\: \underline{ \boxed{\red{ \frak {Work \: done = 19 \: J}}}}$

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