A student places her 442 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by 4.44 cm, then releases the book. What is the book's speed as it slides away? The spring constant is 1102 N/m.
(1/2)*K*x^2 = (1/2)*m*v^2
K-Spring constant, 1102 N/m
x-Compression distance, 4.44cm = .0444m
m-Mass of physics book, 442 g = .442 Kg
Re-arrange:
v = x*(K/m)^0.5
v = .0444*(1102/.442)^0.5
v = 2.22 m/s
To determine the book's speed as it slides away, we can use the principles of energy conservation.
The potential energy stored in the spring when it is compressed is given by the formula:
Potential energy = 1/2 * k * x^2
Where:
k is the spring constant (1102 N/m)
x is the displacement from the equilibrium position (4.44 cm or 0.0444 m)
Substituting the values into the formula, we get:
Potential energy = 1/2 * 1102 N/m * (0.0444 m)^2 = 1/2 * 1102 N/m * 0.0019736 m^2 = 1.083 J
According to the law of conservation of energy, this potential energy gets converted into kinetic energy as the book slides away.
Kinetic energy = 1/2 * m * v^2
Where:
m is the mass of the book (442 g or 0.442 kg)
v is the velocity of the book
To find the velocity, we need to equate the potential energy to the kinetic energy:
1.083 J = 1/2 * 0.442 kg * v^2
Solving for v, we get:
v^2 = (2 * 1.083 J) / 0.442 kg
v^2 = 4.907 J/kg
Taking the square root of both sides, we find:
v ≈ 2.214 m/s
Therefore, the book's speed as it slides away is approximately 2.214 m/s.