Tag: physics

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Multiple choice physics energy and its forms introduction to work work introduction to work and energy

A small ball bearing is releases at the top of a long vertical column of glycerine of height $2h$. The ball bearing falls through a height $h$ in a time $t _{1}$ and then the remaining height with the terminal velocity in time $t _{2}$ Let $W _{1}$ and $W _{2}$ be the work done against viscous drag over these height. therefore.

  1. $t _{1}< t _{2}$

  2. $t _{1}> t _{2}$

  3. $W _{1}=W _{2}$

  4. $W _{1}< W _{2}$
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Multiple choice physics energy and its forms introduction to work work introduction to work and energy

No work is done by a force on an object if 

  1. the force is always perpendicular to its velocity

  2. the force is always perpendicular to its acceleration

  3. the object is stationary but the point of application of the force moves on the object

  4. the object moves in such a way that the point of application of the force remains fixed.
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A Correct answer
Explanation
The work done by a force F in displacing a particle through d is:
$W=\vec{F}\cdot \vec{d}=Fd\cos \theta$
If $\theta =90^o$
$W=Fd\cos 90^o=0$
Now direction of displacement is same as the direction of velocity.
So if force is always perpendicular to the velocity then no work is done.
Multiple choice physics energy and its forms introduction to work work introduction to work and energy

A moving particle is acted upon by several forces $F _1, F _2, F _3 .....$ etc. One of the force is chosen, say $F _2$,  then which of the following statement about $F _2$ will be true.

  1. Work done by $F _2$ will be negative if speed of the particle decreases

  2. Work done by $F _2$ will be positive if speed of the particle increases

  3. Work done by $F _2$ will be equal to the work done by other forces if speed of the particle does not change

  4. If $F _2$ is a conservative force, then work done by all other forces will be equal to change in potential energy due to force $F _2$ when speed remains constant.
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Multiple choice physics energy and its forms introduction to work work introduction to work and energy

Rahul took a wooden cube of volume $1000  {cm}^3$ and put it in water. He observed that $\displaystyle \frac{3}{5}th$ of its volume is below the level of water. Later, he floated the cube in a liquid of density $0.8  g  {cm}^{-3}$ and applied extra force on the cube to completely submerge it in the given liquid. Calculate how much extra force Rahul applied on the cube.

  1. 2 N

  2. 8 N

  3. 6 N

  4. 4 N
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Explanation

Volume = 1000 cm^3. In water (density 1), 3/5 submerged means weight = 0.6 * 1000 * 1 = 600g = 6N. In liquid (density 0.8), buoyant force when fully submerged = 1000 * 0.8 = 800g = 8N. Since weight is 6N, the extra force required to submerge it is 8N - 6N = 2N.