Question

A spring is hanging from the ceiling. Attaching a 700 g physics book to the spring causes it to stretch 24 cm in order to come to equilibrium. What is the book's maximum speed? Express your answer with the appropriate units.

Answer 1

v ≈ 1.93 m/s

Step-by-step explanation:

The potential energy can be calculated as:

PE = (1/2)k(0.24)²

The potential energy is also equal to the kinetic energy (KE) of the book at its maximum speed. So we can equate the two:

PE = KE

Solving for KE:

KE = (1/2)k(0.24)²

Now we need to find the spring constant (k). The spring constant represents the stiffness of the spring and can be determined using Hooke's law:

k = (mg)/x

Where m is the mass of the book and g is the acceleration due to gravity (approximately 9.8 m/s²).

In this case, the mass of the book is 700 g, which is equivalent to 0.7 kg.

Substituting the values into the equation:

k = (0.7 * 9.8) / 0.24

Now we can substitute the value of k back into the equation for KE:

KE = (1/2)(0.7 * 9.8 / 0.24)(0.24)²

Simplifying:

KE = (1/2)(0.7 * 9.8)(0.24)

Finally, we can calculate the value:

KE ≈ 0.82 Joules

The maximum speed (v) of the book can be calculated using the equation:

KE = (1/2)mv²

Solving for v:

v = √(2KE / m)

Substituting the values:

v = √(2 * 0.82 / 0.7)

Calculating the value:

v ≈ 1.93 m/s