
A girl is carrying a school bag of 3 kg mass on her bag and moves 200 meters on a levelled road. The work done against the gravitational force will be \[\left( g=10m/{{s}^{2}} \right)\]
\[\begin{align}
& \text{A}\text{. 6}\times \text{1}{{\text{0}}^{3}}J \\
& \text{B}\text{. 6}J \\
& \text{C}\text{. 0}\text{.6}J \\
& \text{D}\text{. zero} \\
\end{align}\]
Answer
567.3k+ views
Hint: In the given question we have asked the work done against the gravitational force. And work done is given as a dot product of force and displacement and we have given the mass and the displacement. But before solving this we will draw the diagram for better understanding. Also the value of g is given as \[g=10m/{{s}^{2}}\].
Formula used: \[W=Fd\cos \theta \]
Complete step by step answer:
According to the question a girl is carrying a bag of mass 3 kg and moves 200m, this 200 meters basically shows us displacement so it will have magnitude and direction. A diagram for the given question is
F shows the gravitational force or gravitational pull whereas d shows the displacement. Now as the girl is moving on a levelled road and she is carrying the bag so the displacement for the bag will be the same as the girl. And we can also conclude that the force F and displacement d are in perpendicular direction as the road is levelled or a straight horizontal plane.
Now the work done is given as dot product of force and displacement i.e.
\[W=Fd\cos \theta \]
Where f is the force, d is displacement and θ is the angle between F and D.
Here the angle between F and d is \[90{}^\circ \]and \[\cos 90{}^\circ =0\], hence according to the above equation work done against the gravitational force in carrying the bag is zero.
So, the correct answer is “Option D”.
Note: As all the required data is given therefore mistakenly we can neglect the angle and calculate the work done. Also, work done is mostly just given as a product of displacement and force taking force and displacement parallel or antiparallel, even if a girl was moving along a slope we would have obtained some value for work done.
Formula used: \[W=Fd\cos \theta \]
Complete step by step answer:
According to the question a girl is carrying a bag of mass 3 kg and moves 200m, this 200 meters basically shows us displacement so it will have magnitude and direction. A diagram for the given question is
F shows the gravitational force or gravitational pull whereas d shows the displacement. Now as the girl is moving on a levelled road and she is carrying the bag so the displacement for the bag will be the same as the girl. And we can also conclude that the force F and displacement d are in perpendicular direction as the road is levelled or a straight horizontal plane.
Now the work done is given as dot product of force and displacement i.e.
\[W=Fd\cos \theta \]
Where f is the force, d is displacement and θ is the angle between F and D.
Here the angle between F and d is \[90{}^\circ \]and \[\cos 90{}^\circ =0\], hence according to the above equation work done against the gravitational force in carrying the bag is zero.
So, the correct answer is “Option D”.
Note: As all the required data is given therefore mistakenly we can neglect the angle and calculate the work done. Also, work done is mostly just given as a product of displacement and force taking force and displacement parallel or antiparallel, even if a girl was moving along a slope we would have obtained some value for work done.
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