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7 days ago

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A 2.50-kg textbook is forced against a horizontal spring of negligible mass and force constant 250 N/m, compressing the spring a

distance of 0.250 m. When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \mu_kμ k = 0.30. Use the work–energy theorem to find how far the text-book moves from its initial position before coming to rest.

1 answer:

ValentinkaMS [1.1K]7 days ago

6 0

Answer:

L = 1.06 m

Explanation:

According to the work-energy theorem, the total work performed by all forces must match the difference in kinetic energy.

This leads us to:

$W_{spring} + W_{friction} = KE_f - KE_i$

Next, we know that

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

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

Since both the initial and final speeds of the book are zero, its kinetic energy at both points is also zero.

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

Given that

k = 250 N/m

x = 0.250 m

m = 2.50 kg

Now we will insert all the known values into the equation.

$\frac{1}{2}(250)(0.250)^2 = 0.30(2.50)(9.81)L$

At this point, we find

$7.8125 = 7.3575L$

$L = 1.06 m$

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