Problem 42 - University Physics with Modern Physics

Question

A \(2.00-\mathrm{kg}\) block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) , compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope \(37.0^{\circ}\) (Fig. 7.30\()\) . (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

Step-by-Step Solution

Answer

(a) The block's speed is 3.11 m/s. (b) It travels 1.64 m up the incline.

Step 1: Determine the Potential Energy in the Spring

The potential energy stored in the spring when compressed can be calculated using the formula for elastic potential energy: \[ PE_{spring} = \frac{1}{2}k x^2 \] Given values are: \( k = 400 \text{ N/m} \) and \( x = 0.220 \text{ m} \). Substitute these values into the equation: \[ PE_{spring} = \frac{1}{2} \times 400 \times (0.220)^2 \] Calculate the result, \( PE_{spring} = 9.68 \text{ J} \).

Step 2: Convert Spring Potential Energy to Kinetic Energy

Once the block is released from the spring, all potential energy is converted into kinetic energy because the surface is frictionless. The kinetic energy is given by: \[ KE = \frac{1}{2} mv^2 \] where \( m = 2.00 \text{ kg} \) is the mass of the block, and \( v \) is its velocity. Set the kinetic energy equal to the spring potential energy: \[ 9.68 = \frac{1}{2} \times 2.00 \times v^2 \] Solve for \( v \): \[ v^2 = \frac{9.68}{1.00} = 9.68 \] \[ v = \sqrt{9.68} = 3.11 \text{ m/s} \].

Step 3: Calculate the Height Reached on the Incline

As the block slides up the incline, kinetic energy is converted to gravitational potential energy given by \( PE_{gravity} = mgh \). Set \( KE = PE_{gravity} \): \[ \frac{1}{2} mv^2 = mgh \] The mass \( m \) cancels out: \[ \frac{1}{2} (3.11)^2 = 9.8h \] Solve for \( h \): \[ h = \frac{9.68}{9.8} = 0.987 \text{ m} \].

Step 4: Find the Distance Traveled Along the Incline

To find the distance \( d \) traveled along the incline, use the relationship between height and distance on an incline: \[ h = d \sin(37^{\circ}) \] Substitute the value of \( h \): \[ 0.987 = d \times \sin(37^{\circ}) \] \[ d = \frac{0.987}{\sin(37^{\circ})} \approx \frac{0.987}{0.6018} \approx 1.64 \text{ m} \].

Key Concepts

Elastic Potential Energy

When a spring is compressed, it stores energy known as elastic potential energy. This type of potential energy can be calculated using the formula:

• \( PE_{spring} = \frac{1}{2} k x^2 \)

where \( k \) is the spring constant, indicating the stiffness of the spring, and \( x \) is the displacement or compression from its equilibrium position.
In our exercise, a spring with a constant \( k \) of 400 N/m is compressed by 0.220 m. Substituting these values, we find that the elastic potential energy stored is approximately 9.68 Joules. This energy will be converted into kinetic energy once the spring is released, assuming no energy loss due to friction. Understanding how springs store and release energy helps explain many real-world mechanical systems.

Kinetic Energy

Kinetic energy is the energy of motion. When our block is released from the spring, the stored elastic potential energy in the spring is converted into kinetic energy.
Kinetic energy is given by:

• \( KE = \frac{1}{2} mv^2 \)

where \( m \) is the mass of the object and \( v \) its velocity.
In our case of a 2.00 kg block, the spring's entire potential energy changes into kinetic energy because the surface is frictionless. Solving \( \frac{1}{2} \times 2.00 \times v^2 = 9.68 \) gives a velocity of approximately 3.11 m/s. Understanding kinetic energy is crucial as it informs how objects move and interact in the absence of external forces.

Inclined Plane

An inclined plane is a flat surface tilted at an angle, used as a simple machine. It reduces the effort needed to elevate objects by extending the distance over which the force is applied.
In our problem, the incline's slope is given as \(37.0^{\circ}\). As the block moves on this incline, kinetic energy is converted into gravitational potential energy, slowing down the block until it momentarily stops.
The relationship between the height \( h \) and the distance \( d \) traveled on the incline uses trigonometry:

• \( h = d \cdot \sin(\theta) \)

where \( \theta \) is the angle of the incline. In our exercise, the block rises 0.987 m up the incline, resulting in a traveled distance of approximately 1.64 m. Understanding the dynamics of inclined planes is essential in solving various physics problems involving slopes and ramps.

Gravitational Potential Energy

As the block moves up the incline, its kinetic energy is turned into gravitational potential energy. This form of energy depends on the object's height above a reference level and is calculated using:

• \( PE_{gravity} = mgh \)

where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
For the moving block, as kinetic energy equals gravitational potential energy when at the peak height, we find \( h \) by using the block's initial kinetic energy of 9.68 Joules: \( h = \frac{9.68}{9.8} \), resulting in approximately 0.987 m.
This principle is foundational in physics, allowing us to predict how high an object will rise given its kinetic energy.