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
[tex]W_{spring} + W_{friction} = KE_f - KE_i[/tex]
now we know that
[tex]W_{spring} = \frac{1}{2}kx^2[/tex]
[tex]W_{friction} = -\mu mg L[/tex]
initial and final speed of the book is zero so initial and final kinetic energy will be zero
[tex]\frac{1}{2}kx^2 - \mu mg L= 0 - 0[/tex]
here we know that
k = 250 N/m
x = 0.250 m
m = 2.50 kg
now plug in all data in it
[tex]\frac{1}{2}(250)(0.250)^2 = 0.30(2.50)(9.81)L[/tex]
now we have
[tex]7.8125 = 7.3575L[/tex]
[tex]L = 1.06 m[/tex]
The distance that the textbook moves from its initial position before coming to rest is 1.0629m
According to the work-energy theorem, the work done on the spring is equal to the work done by friction
Work done by the spring = [tex]\frac{1}{2}kx^2[/tex]
Work done by friction = [tex]\mu mgL[/tex]
k is the spring constant = 250 N/m
x is the distance moved by the spring = 0.250m
[tex]\mu[/tex] is the coefficient of friction = 0.30
m is the mass of the textbook = 2.50kg
g is the acceleration due to gravity = 9.8m/s²
L is the distance from its initial position before coming to rest
Using the formula
[tex]\frac{1}{2} kx^2=\mu mgL[/tex]
Substitute the given parameters into the formula as shown;
[tex]\frac{1}{2} (250)(0.25)^2=0.3 (2.5)(9.8)L\\7.8125=7.35L\\L=\frac{7.8125}{7.35}\\L= 1.0629m[/tex]
Hence the distance that the textbook moves from its initial position before coming to rest is 1.0629m
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