Problem 63 - University Physics with Modern Physics

Question

A 0.150-kg block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 m above the floor. The spring has force constant 1900 N/m and is initially compressed 0.045 m. The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

Step-by-Step Solution

Answer

The speed of the block when it reaches the floor is 7.01 m/s.

Step 1: Calculate Potential Energy in Compressed Spring

The initial potential energy in the spring when compressed can be calculated using the formula for elastic potential energy: \[ PE_{spring} = \frac{1}{2} k x^2 \] where \( k = 1900 \, \text{N/m} \) is the spring constant and \( x = 0.045 \, \text{m} \) is the compression distance. Plug in the values: \[ PE_{spring} = \frac{1}{2} \times 1900 \times (0.045)^2 \] \[ PE_{spring} = 1.92 \, \text{J} \].

Step 2: Calculate Kinetic Energy as Block Leaves the Table

Assuming negligible friction, the potential energy stored in the spring is converted entirely into the kinetic energy of the block as it leaves the table. Thus, we can set \( KE = PE_{spring} \). Using the kinetic energy formula: \[ KE = \frac{1}{2} m v^2 \] where \( m = 0.150 \, \text{kg} \) is the mass of the ice block. Equate and solve for \( v \): \[ \frac{1}{2} \times 0.150 \times v^2 = 1.92 \] \[ v^2 = \frac{1.92}{0.075} \] \[ v^2 = 25.6 \] \[ v = 5.06 \, \text{m/s} \].

Step 3: Calculate Velocity of the Block Upon Reaching the Floor

As the block falls from the table, its velocity increases due to gravitational acceleration. Calculate the additional speed gained using: \[ v_f^2 = v^2 + 2gh \] where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h = 1.20 \, \text{m} \) is the height of the table. Plug in the values: \[ v_f^2 = (5.06)^2 + 2 \times 9.81 \times 1.20 \] \[ v_f^2 = 25.6 + 23.544 \] \[ v_f^2 = 49.144 \] \[ v_f = 7.01 \, \text{m/s} \].

Key Concepts

Potential Energy

Potential energy is the stored energy of an object due to its position or configuration. In the context of springs, we talk about elastic potential energy. Elastic potential energy is stored in a compressed or extended spring, ready to be converted into kinetic energy. For a compressed or stretched spring, the potential energy can be calculated using the formula:
\[PE_{spring} = \frac{1}{2} k x^2 \]where:

• \( k \) is the spring constant, a measure of the stiffness of the spring.
• \( x \) is the distance the spring is compressed or extended.

In our problem, the spring has a constant \( k = 1900 \, \text{N/m} \) and is compressed \( 0.045 \, \text{m} \). Doing the math gives us the potential energy stored, \( PE_{spring} = 1.92 \, \text{J} \). This energy is crucial as it later transforms into kinetic energy when released.

Kinetic Energy

Kinetic energy is the energy of motion. Once the potential energy in a spring is released, it is largely converted into the kinetic energy of the moving object. The formula for kinetic energy is:
\[KE = \frac{1}{2} m v^2\]where:

• \( m \) is the mass of the object.
• \( v \) is the velocity of the object.

In the scenario given, the potential energy of the spring is transformed into the kinetic energy of the ice block as it leaves the table. By assuming negligible friction, we make the calculation straightforward by setting \( KE = PE_{spring} \). This allows us to solve for velocity \( v \) of the block, which turns out to be \( 5.06 \, \text{m/s} \). This velocity is before it starts gaining additional speed due to gravity.

Gravitational Acceleration

Gravitational acceleration is the increase in speed of an object as it is pulled towards the Earth. When the ice block leaves the edge of the table, it is subject to the force of gravity, with an acceleration of \( g = 9.81 \, \text{m/s}^2 \). This force causes the block to gain velocity as it descends.The effect of gravity on falling objects can be mathematically expressed through the kinematic equation:
\[v_f^2 = v^2 + 2gh\]where:

• \( v \) is the initial velocity when the block leaves the table.
• \( g \) is the gravitational acceleration.
• \( h \) is the height from which the object falls.

For the block of ice, it starts with an initial velocity \( v = 5.06 \, \text{m/s} \) from the spring release and gains additional speed as it falls from a height of \( 1.20 \, \text{m} \). Solving the equation reveals its impact speed as \( 7.01 \, \text{m/s} \). This showcases gravitational potential energy converting to kinetic energy as it falls.

Spring Mechanics

Spring mechanics involve the principles of force and energy as applied to springs. Springs store energy when compressed or stretched and release this energy as they return to their original form. The two main types of energy involved in spring mechanics are potential energy and kinetic energy.The spring constant \( k \) is fundamental in determining how much a spring will compress or extend under a given force. It is essentially a measure of stiffness:

• A higher \( k \) means a stiffer spring.
• A lower \( k \) means a more flexible spring.

The relation between force, spring constant, and displacement is expressed by Hooke's Law:
\[F = -kx\]where:

• \( F \) is the force applied.
• \( x \) is the displacement from equilibrium.

In our problem, once the spring is released, it transfers potential energy into kinetic energy, which propels the ice block. This conversion highlights the efficiency of springs in energy transformation. Understanding spring mechanics helps us appreciate how everyday items like mattresses and car suspensions work effectively.