A) \[1\,\,unit\]
B) \[20\,\,unit\]
C) \[5\,\,unit\]
D) \[zero\]
Correct Answer: A
Solution :
Work \[(W)\] is measured by the vector product of the applied force \[(\overset{\to }{\mathop{\mathbf{F}}}\,)\] and the displacement \[(\overset{\to }{\mathop{\mathbf{s}}}\,)\] of the body in the direction of the force\[(\overset{\to }{\mathop{\mathbf{F}}}\,)\], that is, \[W=\overset{\to }{\mathop{\mathbf{F}}}\,\cdot \overset{\to }{\mathop{\mathbf{s}}}\,\] Given,\[\overset{\to }{\mathop{\mathbf{F}}}\,=2\widehat{\mathbf{i}}+3\widehat{\mathbf{j}}-5\widehat{\mathbf{k}},\,\,\overset{\to }{\mathop{\mathbf{s}}}\,=2\widehat{\mathbf{i}}+4\widehat{\mathbf{j}}+3\widehat{\mathbf{k}}\] \[\therefore \] \[W=(2\widehat{\mathbf{i}}+3\widehat{\mathbf{j}}-5\widehat{\mathbf{k}})\cdot (2\widehat{\mathbf{i}}+4\widehat{\mathbf{j}}+3\widehat{\mathbf{k}})\] \[=(2)(2)+(3)(4)+(-5)(3)\] \[=4+12-15\] \[=1\,\,unit\] Note: Both force and displacement are vector quantities but work is a scalar quantity.
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