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BlackZzzverrR [31]
3 years ago
8

If you are holding a book and you let go of it, why does it fall? Explain in terms of forces, potential energy, and kinetic ener

gy.
Physics
1 answer:
inessss [21]3 years ago
3 0

Answer:

Let the mass of the book be "m", acceleration due to gravity be "g", velocity be "v" and height be "h".

Now if we are holding a book at a certain height (h), <em><u>the potential energy will be maximum which is equal to mass× acceleration due to gravity× height (= mgh)</u>.</em>

(Remember: kinetic energy =0)

Now we consider that the book is dropped, in this case a force will act downward towards the centre of the earth, <em><u>Force= mass× acceleration due to gravity (F=mg)</u></em>. It is equal to the weight of the book.

While the book is falling, the potential energy stored in the book converts into kinetic energy and strikes the floor with <em><u>the maximum kinetic energy= (1/2)×mass×velocity² (=1/2mv²)</u>.</em>

(Remember: kinetic energy=0)

Due to this process the whole energy is conserved.

As the potential energy decreases kinetic energy increases.

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A 65 N boy sits on a sled weighing 52 N on a horizontal surface. The coefficient of friction between the sled and the snow is 0.
pogonyaev

Answer:

1.40 N

Explanation:

The magnitude of the frictional force is given by:

F=\mu N

where

\mu is the coefficient of friction

N is the magnitude of the normal reaction

The coefficient of friction for this problem is \mu=0.012. The magnitude of the normal reaction is equal to the combined weight of the boy and the sled, because the surface is horizontal, so

N=65 N+52 N=117 N

Therefore, the frictional force is

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Answer:

The distance covered by puck A before collision is  z = 8.56 \ m

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  v_A =  3.90 \ m/s

        The speed of puck B is  v_B  =  4.30 \ m/s

The distance covered by puck A is mathematically represented as

     z =  v_A * t

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 The distance covered by puck B  is  mathematically represented as

      18 - z =  v_B  * t

=>   t  = \frac{18 - z}{v_B}

Since the time take before collision is the same

        \frac{18 - z}{V_B}  =  \frac{z}{v_A}

substituting values

          \frac{18 -z }{4.3}  = \frac{z}{3.90}

=>      70.2 - 3.90 z   = 4.3 z

=>       z = 8.56 \ m

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Answer:

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