Question

a car of mass 1500 kg is dropped by parachute from a helicopter, hits the ground at a speed of 4.5 m/s, and lands on all four tires. each tire has a spring whose spring constant is 3.1 x 10⁵ n/m. how much will the springs compress?

Answer 1

The springs will compress approximately 31.3 cm when the car with a mass of 1500 kg hits the ground at a velocity of 4.5 m/s.

Step-by-step explanation:

To find out how much the springs will compress when the 1500 kg car lands on them, we can use the energy conservation principle. The potential energy at the maximum compression of the springs will be equal to the kinetic energy of the car just before it hits the ground.

The kinetic energy (KE) of the car when it touches the ground is given by:

KE = ½ * m * v²

where:

• m = mass of the car = 1500 kg
• v = velocity of the car = 4.5 m/s

The potential energy (PE) stored in a spring is given by:

PE = ½ * k * x²

where:

• k = spring constant = 3.1 x 10⁵ N/m
• x = compression distance

Setting KE equal to PE to find x:

½ * m * v² = ½ * k * x²

Upon solving for x, we get the expression:

x = √((m * v²) / k)

Plugging in the numbers:

x = √((1500 * 4.5²) / (3.1 x 10⁵))

x ≈ √((1500 * 20.25) / (3.1 x 10⁵))

x ≈ √((30375) / (3.1 x 10⁵))

x ≈ √(0.098) m

x ≈ 0.313 m or 31.3 cm

Each spring will compress by approximately 31.3 cm upon impact.