Problem 6 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 1200 -kg car rolling on a horizontal surface has speed \(v=75 \mathrm{~km} / \mathrm{h}\) when it strikes a horizontal coiled spring and is brought to rest in a distance of \(2.2 \mathrm{~m}\). What is the spring stiffness constant of the spring?

Step-by-Step Solution

Answer

The spring stiffness constant is approximately 107644.85 N/m.

Step 1: Convert Velocity to m/s

First, convert the velocity from km/h to m/s. The car velocity is given as \( v = 75 \mathrm{~km/h} \). To convert km/h to m/s, we multiply by \( \frac{1000}{3600} \). So the velocity in meters per second is \( v = 75 \times \frac{1000}{3600} = 20.83 \mathrm{~m/s} \).

Step 2: Calculate Initial Kinetic Energy

The initial kinetic energy \( KE \) of the car can be calculated using the formula \( KE = \frac{1}{2} m v^2 \) where \( m = 1200 \mathrm{~kg} \) and \( v = 20.83 \mathrm{~m/s} \). Plugging in these values, \( KE = \frac{1}{2} \times 1200 \times (20.83)^2 = 260416.66 \mathrm{~J} \).

Step 3: Understand Energy Conservation

When the car is brought to rest, the kinetic energy is converted into potential energy stored in the spring. This means \( KE = PE_{spring} \), where \( PE_{spring} = \frac{1}{2} k x^2 \), and \( k \) is the spring constant we need to find, and \( x = 2.2 \mathrm{~m} \) is the compression of the spring.

Step 4: Solve for Spring Constant

Using the equation \( KE = \frac{1}{2} k x^2 \) and the values we have, substitute \( 260416.66 = \frac{1}{2} k (2.2)^2 \). Solving for \( k \), we multiply both sides by 2 to get \( 520833.32 = k \times 4.84 \). Divide both sides by 4.84 to find \( k \):\[ k = \frac{520833.32}{4.84} \approx 107644.85 \mathrm{~N/m}. \]

Key Concepts

Kinetic Energy of Moving Objects

Kinetic energy is the energy that an object possesses due to its motion. It's calculated using the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object in kilograms, and \( v \) is the velocity of the object in meters per second.

• For our exercise, the car has a mass \( m = 1200 \) kg and a speed \( v = 75 \) km/h.
• Converting the speed to meters per second, we get \( v = 20.83 \) m/s.
• Substituting these values into the formula gives us the initial kinetic energy: \( KE = 260416.66 \) J.

The car's kinetic energy before hitting the spring is what determines the energy transferred to the spring upon impact.

Energy Conservation in Action

Energy conservation is a fundamental principle in physics stating that energy cannot be created or destroyed, only transformed from one form to another. In this exercise, we see the conservation of energy in a tangible way.

• The car's initial kinetic energy is entirely transformed into potential energy when it compresses the spring.
• This conversion follows the equation \( KE = PE_{spring} \) where the kinetic energy of the car becomes the potential energy stored in the spring.

This principle allows us to solve for unknown quantities, such as the spring stiffness constant \( k \), which is a measure of how much force is needed to compress the spring by a unit length.

Potential Energy in Springs

Potential energy in a compressed or stretched spring can be calculated using the formula \( PE_{spring} = \frac{1}{2} k x^2 \), where \( k \) is the spring constant, and \( x \) is the compression or extension distance of the spring.

• The spring constant \( k \) tells us how stiff the spring is – a larger \( k \) means a stiffer spring.
• In our situation, the car compresses the spring by \( x = 2.2 \) m.
• Using the energy conservation principle \( KE = PE_{spring} \) and substituting known values, we solve for \( k \) using the equation: \( \frac{1}{2} k x^2 = 260416.66 \) J.
• This turns into an equation for finding \( k \), resulting in \( k \approx 107644.85 \) N/m.

Understanding potential energy in springs provides insights into how energy is stored and later released, reflecting the characteristics of the spring itself.