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Cerrena [4.2K]
3 years ago
8

A 39-cm-long vertical spring has one end fixed on the floor. Placing a 2.2 kg physics textbook on the spring compresses it to a

length of 29 cm.What is the spring constant?
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

To solve this problem we will apply Hooke's law. Which says that the force on a spring is equivalent to the product between the elastic constant and the displacement of the spring.

F = k x

Here,

k = Spring constant

x = Displacement

Rearranging to find the spring constant,

k = \frac{F}{x}

Spring compresses it to 29 cm so

x = 39- 29 = 10 cm

k = \frac{mg}{x} \rightarrow The force in this case is equivalent to the weight of the spring

k = \frac{2.2 * 9.8}{10 * 10 ^ {-2}}

k = 215.6N/m

Therefore the spring constant is 215.6N/m

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Which if, any, of these statements are true? (More than one may be true.) Assume the batteries are ideal. Check all that apply.
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3 years ago
A 50.3-kg person, running horizontally with a velocity of 2.44 m/s, jumps onto a 13.4-kg sled that is initially at rest. (a) Ign
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Answer:

a

  v =  1.9267 \  m/s

b

The value is    \mu = 0.0063        

Explanation:

From the question we are told that

      The  mass of the person is  m_1  =  50.3 \  kg

       The  horizontal velocity is  u_1  =  2.44 \  m/s

       The mass of the shed is  m_2  =  13.4 \  kg

        The distance covered is  d =  30 \ m

Generally from the law of momentum conservation we have that

        m_1 *  u_1 + m_2 * u_2 =  (m_1 + m_2)v

Here  u_2 is the initial velocity of the shed which is  0 m/s  

       50.3 *  2.44 + 13.4  * 0 =  (50.3 + 13.4) v

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Generally the workdone by friction is mathematically represented as  

          W = \Delta  KE          

          W =  \frac{1}{2}  *   m  *  (v_f  -  v  )

Here v_f is the final velocity of the person and the shed when they come to rest and the value is  v_f = 0 \ m/s

 Generally this workdone  by friction is also mathematically represented as

           W = - \mu *  m *  g * d

=>        - \mu *  m *  g * d =  \frac{1}{2}  *   m  *  (v_f  -  v  )

=>       \mu =  - \frac{0.5 *  ( v_f^2 - v^2 )}{g * d }

=>         \mu =-  \frac{0.5 *  ( 0^2 -  1.9267^2 )}{9.8 *  30 }

=>         \mu = 0.0063        

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