Answer:
0.123 m.
Explanation:
From Hook's law,
The potential energy of the book = the energy stored in the spring.
mgh = 1/2ke².................. Equation 1
Where m= mass of the book, g = acceleration due to gravity, h = height, k = spring constant of the spring, e = distance of compression.
make e the subject of the equation
e = √(2mgh/k).................. Equation 2
Given: m = 1.3 kg, h = 0.8 m, k = 1350 N/m
Constant: g = 9.8 m/s²
Substitute into equation 2
e = √(2×1.3×0.8×9.8/1350)
e = √(20.384/1350)
e = √(0.0151)
e = 0.123 m.
0.015m (downwards)
When the book is dropped on the top of the spring at that height, the potential energy ([tex]E_{P}[/tex]) of the book is converted to elastic energy ([tex]E_{E}[/tex]) on the spring thereby causing a compression on the spring. i.e
[tex]E_{P}[/tex] = [tex]E_{E}[/tex]
But;
The potential energy [tex]E_{P}[/tex] of the mass (book), is the product of the mass(m) of the book, the height(h) from which it was dropped and the acceleration due to gravity (g). i.e
[tex]E_{P}[/tex] = - m x g x h [the -ve sign shows a decrease in height as the mass (book) drops]
Also;
The elastic energy ([tex]E_{E}[/tex]) of compression of the spring is given by
[tex]E_{E}[/tex] = [tex]\frac{1}{2}[/tex] x k x c
Where;
c = compression length of the spring
k = the spring's constant
Substitute these values of [tex]E_{P}[/tex] and E into equation (i) as follows;
- m x g x h = [tex]\frac{1}{2}[/tex] x k x c ----------------(ii)
From the question;
m = 1.3kg
h = 0.8m
Take g = 10m/s²
k = 1350N/m
Substitute these values into equation (ii) as follows;
- 1.3 x 10 x 0.8 = [tex]\frac{1}{2}[/tex] x 1350 x c
- 10.4 = 675c
Solve for c;
c = - 0.015 m [The negative sign shows that the spring actually compresses]
Therefore, the maximum distance the spring will be compressed is 0.015m (downwards of course).
