A mass (m = 30 g) falls onto a spring (k = 7.3 N/m) from a height (h = 25 cm). The spring compresses an additional amount x before temporarily coming to a stop. What is the value of x?

x₁ = 0.1878 m

Explanation:

For this exercise we will use conservation of energy

starting point. Highest point

         Em₀ = U = m g h

final point. Lowest point with fully compressed spring

         Em_f = K_e + U

         Em_f = ½ K x² + m g x

energy is conserved

         Em₀ = Em_f

        m g h = ½ K x² + m g x

       ½ K x² + mg (x- h) = 0

let's substitute

       ½ 7.3 x² + 0.030 9.8 (x- 0.25) = 0

        3.65 x² + 0.294 (x- 0.25) = 0

        x² + 0.080548 (x- 0.25) = 0

        x² - 0.020137 + 0.080548 x = 0

        x² + 0.080548 x - 0.020137 = 0

let's solve the quadratic equation

      x = [0.080548 ±√ (0.080548² + 4   0.020137)] / 2

      x = [0.080548 ± 0.29502] / 2

      x₁ = 0.1878 m

      x₂ = -0.1072 m

These are the compression and extension displacement of the spring

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm