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  • A body of mass 5 kg is placed at the origin, and can move only on the x-axis. A force of 10 N is acting on it in a direction making an angle of 60 with the x-axis and displaces it along the x-axis

A body of mass 5 kg is placed at the origin, and can move only on the x-axis. A force of 10 N is acting on it in a direction making an angle of 60 with the x-axis and displaces it along the x-axis by 4 metres. The work (in joule) done by the force is 

Option: 1 2.5

Option: 2 7.25

Option: 3 40

Option: 4 20

Answers (1)

best_answer

 

 

Nature of Work Done -

 

  1. Positive Work-

  • Positive work means that force (or its component) is parallel to displacement.

  • Means 0\leq \Theta < \frac{\pi }{2}

                     Where \Theta is the angle between force vectors and displacement vector

  • Maximum work = W_{max} = FS,\ When\ \theta = 0^{0}

  • E.g When you move a block by pulling it then work done by you on the block is positive

 

  1. Negative Work

  • Negative work means that force (or its component) is opposite to displacement.

  • Means \frac{\pi }{2}< \Theta \leq \pi

                     Where \Theta is the angle between force vectors and displacement vector

  • Minimum work= W_{min} = -FS,\ When\ \theta = 180^{0}

  • E.g When a body is made to slide over a rough surface, the work done by the frictional force is negative

 

  1. Zero work

  • Under three condition Work can be zero

             a. If the force is perpendicular to the displacement

                      Means \Theta = \frac{\pi }{2}

                 E.g-When a body moves in a circle the work done by the centripetal force is always zero.

             b. If there is no displacement (means s = 0)

                 E.g- When a person tries to displace a wall by applying a force and can't able to move the wall

                        So the work done by the person on the wall is zero.

             c. If there is no force acting on the body (means F=0)

                 E.g-Motion of an isolated body in free space.

-

So, for any angle,

 

W = \vec{F}.\vec{S}=FScos\theta = 10\times 4\times cos60 ^{\circ} = 10 \times 4 \times \frac{1}{2} = 20 J

Posted by

manish painkra

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