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A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressing the spring 0.123 m. After the block is released, the block moves 0.281 m to the right before coming to rest. The acceleration of gravity is 9.81 m/s 2 . What is the coefficient of kinetic friction between the horizontal surface and the block?

Respuesta :

Answer:

[tex]U_k = 0.113[/tex]

Explanation:

using the law of the conservation of energy:

[tex]E_i -E_f=W_f[/tex]

[tex]\frac{1}{2}Kx^2=NU_kd[/tex]

where K is the spring constant, x is the spring compression, N is the normal force of the block, [tex]U_k[/tex] is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

[tex]\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)[/tex]

solving for [tex]U_k[/tex]:

[tex]U_k = 0.113[/tex]

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