Question

A trolly of mass 1kg is connected to two identical spring of spring constant 100n/ m each . If trolly is displaced from position by 1cm then maximum velocity of trolley is

Answer 1

Answer:

the answer is explained above

Answer 2

Maximum velocity of trolley is 0.14 $(\frac{m}{s})$.

Explanation:

1.Mass of object =1 kg.

 Spring stiffness of each spring =100$(\frac{N}{m})$.

 Object is displaced from its position = 1 cm = 0.01 (m)

2. Here are two identical spring, but it is not given that it is connected in series or parallel. We can take series connection of spring.

  so total spring stiffness (k)=100+100 = 200 $(\frac{N}{m})$

3. When spring is displaced from its neutral position, energy stored in form of potential energy.

  Potential energy of spring (U) = $\frac{1}{2}\times k\times x^{2}$  ...1)

4. When spring come back to its neutral position its total potential energy converted into kinetic energy which is maximum.

 Kinetic Energy(K.E) =$\frac{1}{2}\times m\times v_{max}^{2}$   ...2)

5. Now from conservation of energy

   Maximum kinetic energy at equilibrium position = potential energy of spring in extension position

 

  from equation 1) and  equation 2)

$\frac{1}{2}\times m\times v_{max}^{2}$=$\frac{1}{2}\times k\times x^{2}$     ...3)

6.  Now equation 3) can be written as  

   $v_{max}= x\times \sqrt{\frac{k}{m}}$

   $v_{max}= 0.01\times \sqrt{\frac{200}{1}}$

   $v_{max}$= 0.14$(\frac{m}{s})$