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3 years ago

8

A block with mass of 0.5 kg is forced against a horizontal spring of negligible mass. compressing the spring a distance of 0.20m

. When released, the block moves on a horizontal tabletop for 1 meter before coming to rest. The spring constant is 100N/m. What is the coefficient of friction between the block and the tabletop??

2 answers:

Alik [6]3 years ago

7 0

Answer:

$\mu = 0.408$

Explanation:

Here we know that work done by all force must be equal to change in kinetic energy of the block

so here we will have

$W_{friction} + W_{spring} = K_f - K_i$

here we know that

$W_{friction} = -\mu mg d$

$W_{spring} = \frac{1}{2}kx^2$

also we know that initially and finally block is at rest

so we have

$\frac{1}{2} kx^2 - \mu mg d = 0 - 0$

$\frac{1}{2}(100)(0.20^2) - \mu (0.5)(9.81)(1) = 0$

$\mu = 0.408$

miv72 [106K]3 years ago

5 0

<span>Required"mu" is given by:

mu*0.5*9.8*1 = 0.5*100*0.20^2 or
mu = 4/9.8 = 0.408 or 0.41

Therefore, </span><span> the coefficient of friction between the block and the tabletop is 0.408 or rounded off to 0.41.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

</span>

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