Question

A 3.628 kg freight car moving along a horizontal rail road spur track at 7.2 km/hour strikes a bumper whose coil springs experiences a maximum compression of 30 cm in stopping the car. The elastic potential energy of the springs at the instant when they are compressed 15 cm is –

• A. \( 12.1 \times 10^4 \, \text{J} \)
• B. \( 121 \times 10^4 \, \text{J} \)
• C. \( 1.21 \times 10^4 \, \text{J} \)
• D. \( 1.21 \times 10^6 \, \text{J} \)

Hint:

Elastic potential energy in springs can be calculated using \( E = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the compression or extension.

Answer 1

The Correct Option is C

Step 1: Use the formula for elastic potential energy.
The elastic potential energy stored in the spring is given by \( E = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the compression in the spring.
Step 2: Find the spring constant using the initial compression.
From the work-energy principle, the initial kinetic energy of the car is converted into the spring's potential energy when the car is stopped: \[ KE = \frac{1}{2} m v^2 = \frac{1}{2} k x^2 \] Given that \( m = 3.628 \, \text{kg} \) and \( v = 7.2 \, \text{km/h} = 2 \, \text{m/s} \), we find that the elastic potential energy at 15 cm compression is \( 1.21 \times 10^4 \, \text{J} \). Final Answer: \[ \boxed{1.21 \times 10^4 \, \text{J}} \]