Answer:
d = 1.875 m
Explanation:
Given that
m = 2 kg
k = 250 N/m
x= 0.3 m
Lets take the speed of book just before when it is leaving the spring
From energy conservation
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2[/tex]
[tex]kx^2=mv^2[/tex]
250 x 0.3²= 2 x v²
v= 3.35
When it moves on the rough surface then de -acceleration a
a= μk g
a= 0.3 x 10 = 3 m/s²
The distance cover by book before getting to rest position is d
The final speed(V) of the block will be zero.
We know that
V² = v²- 2 a d
0 = 125 x 0.3² - 2 x 3 x d
d = 1.875 m