a 1.2 kg book is dropped from a height of 0.8 m onto a spring with force constant k=1960 N/m ang neglible mass. find the maximum distance that the spring will be compressed.
mgh=kx^2/2,
x=sqrt(2mgh/k)
To find the maximum distance that the spring will be compressed, we need to determine the potential energy stored in the spring when it reaches its maximum compression.
The potential energy stored in a spring is given by the equation:
PE = (1/2) * k * x^2
Where:
PE is the potential energy stored in the spring,
k is the force constant of the spring, and
x is the compression or elongation of the spring.
In this case, the potential energy is equal to the gravitational potential energy of the book when it is dropped from a height of 0.8 m.
Gravitational potential energy is given by the equation:
PE = m * g * h
Where:
m is the mass of the book,
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
h is the height from which the book is dropped.
Given that the mass of the book is 1.2 kg and the height is 0.8 m, we can calculate the potential energy as:
PE = 1.2 kg * 9.8 m/s^2 * 0.8 m
= 9.408 J
Since the potential energy stored in the spring is equal to the potential energy of the book, we can equate the two equations:
(1/2) * k * x^2 = PE
Substituting the values, we get:
(1/2) * 1960 N/m * x^2 = 9.408 J
Simplifying the equation:
980 N/m * x^2 = 9.408 J
x^2 = 0.0096 m^2
Taking the square root of both sides, we get:
x = 0.098 m
Therefore, the maximum distance that the spring will be compressed is approximately 0.098 m.