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Vinil7 [7]
3 years ago
5

A 2 kg book is pushed from rest to a final velocity of 3 m/s. The book travels 2 m. How much force was the book pushed with

Physics
2 answers:
Studentka2010 [4]3 years ago
6 0

Explanation:

first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s

now using the 3rd law of motion

v^2=u^2+2as

0=9+2a2

a= -9/4m/s^2

now force=mass×accelration

=2kg×(-9/4)m/s^2

=4.5 N

4.5 newton force applied on the book!

✌️:)

Natali5045456 [20]3 years ago
6 0

Answer:

4.5N

Explanation:

Step 1:

Data obtained from the question:

Mass (m) = 2Kg

Initial velocity (U) = 0

Final Velocity (V) = 3m/s

Distance travelled (s) = 2m

Force (F) =?

Step 2:

Determination of the acceleration of the book. This is illustrated below:

Using V^2 = U^2 + 2as, the acceleration (a) can be obtained as follow:

V^2 = U^2 + 2as

3^2 = 0 + 2 x a x 2

9 = 4a

Divide both side by the coefficient of a i.e 4

a = 9/4

a = 2.25m/s2

Step 3:

Determination of the force applied. This is illustrated below:

Force = Mass x Acceleration

F = m x a

F = 2 x 2.25

F = 4.5N

Therefore, a force of 4.5N was used to push the book.

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substituting this into the equation

      KE  =  \frac{ 1}{2} *\frac{GMm}{r}

Now the gravitational force of the planet is mathematically represented as

      F_g  = \frac{GMm}{r^2}

Where M is the mass of the planet and  m is the mass of the satellite

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     KE  =  \frac{ 1}{2} *[\frac{GMm}{r^2}] * r

=>    KE  =  \frac{ 1}{2} *F_g * r

substituting values

       KE  =  \frac{ 1}{2} *6600 * 2.3*10^{7}

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From the question we are told that

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              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

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               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

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                P  = \frac{1}{f}

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Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

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This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

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