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mrs_skeptik [129]
3 years ago
8

A block of mass m = 0.13 kg is set against a spring with a spring constant of k1 = 623 N/m which has been compressed by a distan

ce of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 487 N/m.(A) How far d2, in meters, will the second spring compress when the block runs into it?(B) How fast v, in meters per second, will the block be moving when it strikes the second spring?(C) Now assume friction is present on the surface in between the ends of the springs at their equilibrium lengths, and the coefficient of kinetic friction is ?k = 0.5. If the distance between the springs is x = 1 m, how far d2, in meters, will the second spring now compress?
Physics
1 answer:
iris [78.8K]3 years ago
8 0

Answer:

a) 0.11 m

b) 6.9 m/s

c) 0.10 m

Explanation:

We need to apply the conservation of energy theorem:

U_e+K_{o}+U_o=K_f+U_f

because the surface is frictionless all the elastic potential energy is converted in kinetic, so:

\frac{1}{2}*k*x^2+0+0=K_f\\\\K_f=3.1J

the second spring received 3.1J of energy, so:

3.1J=\frac{1}{2}*487N/m*(x_2)^2\\x_2=0.11m

We know the value of the kinetic energy so:

\frac{1}{2}m*(v_2)^2=3.1J\\v_2=\sqrt{\frac{3.1J*2}{0.13kg}}\\v_2=6.9m/s

now if the surface has friction:

3.1J-\µ*m*g*d=K_f\\3.1J-0.63J=K_f\\K_f=2.5J\\\\2.5J=\frac{1}{2}*K*x^2\\x=\sqrt{\frac{2*2.5J}{487}}\\x=0.10m

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The angle between the blue beam and the red beam in the acrylic block is  

 \theta _d  =0.19 ^o

Explanation:

From the question we are told that

     The  refractive index of the transparent acrylic plastic for blue light is  n_F  =  1.497

     The  wavelength of the blue light is F  =  486.1 nm  =  486.1 *10^{-9} \ m

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       The  wavelength of the red light is C =  656.3 nm  = 656.3 *10^{-9} \  m

    The incidence angle is  i  =  45^o

Generally from Snell's law the angle of refraction of the blue light  in the acrylic block  is mathematically represented as

       r_F =  sin ^{-1}[\frac{sin(i) *  n_a }{n_F} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_F =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.497} ]

      r_F  =  28.18^o

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

       r_C =  sin ^{-1}[\frac{sin(i) *  n_a }{n_C} ]

Where  n_a is the refractive index of air which have a value ofn_a =  1

So

     r_C =  sin ^{-1}[\frac{sin(45) *  1 }{ 1.488} ]

      r_F  =  28.37^o

The angle between the blue beam and the red beam in the acrylic block

     \theta _d  =  r_C  - r_F

substituting values

       \theta _d  = 28.37 -  28.18

       \theta _d  =0.19 ^o

 

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