Question

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.

Answer 1

Answer:

L = 1.06 m

Explanation:

As per work energy theorem we know that work done by all forces must be equal to change in kinetic energy

so here we will have

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

now we know that

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

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

initial and final speed of the book is zero so initial and final kinetic energy will be zero

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

here we know that

k = 250 N/m

x = 0.250 m

m = 2.50 kg

now plug in all data in it

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

now we have

$7.8125 = 7.3575L$

$L = 1.06 m$