Question

A brick of mass ml = 0.11 kg is set against a spring with a spring constant of ki 511 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 228 N/m Otheexpertta.com 33%

Part (a) How far, d2 in meters, will the second spring compress when the brick runs into it? 33%
Part (b) How fast, v in meters per second, will the brick be moving when it strikes the second spring? 33%
Part (c) Now assume that the surface is rough (that is not frictionless). You perform the experiment and observe that the second spring only compresses a distance d2/2. How much energy, in joules, was lost to friction?

Answer 1

Answer:

a)   x₂ = 0.1497 m, b) v = 4,819 m / s, c) W = 1.916 J

Explanation:

a) For this part we use the concept of mechanical energy,

Starting point Maximum compression of spring 1

            Em₀ = $K_{e}$ = ½ k₁ x₁²

Final point. Maximum compression of spring 2

            $Em_{f}$ = $K_{e}$ Ke = ½ k₂ x₂²

As there is no rubbing the energy is conserved

            Em₀ =$Em_{f}$

            ½ k₁ x₁² = ½ k₂ x₂²

            x₂ = x₁ √ k₁ / k₂

            x₂ = 0.1 √ 511/228

            x₂ = 0.1497 m

b) Let's use energy conservation,

Final point. When the spring has no compression

           $Em_{f}$  = K = ½ m v²

          Em₀ = $Em_{f}$  

          ½ k₁ x₁² = ½ m v²

          v = x₁ √ k₁ / 2m

          v = 0.1 √ (511/2 0.11)

          v = 4,819 m / s

c) If the surface is rough, some of the energy is lost in heat, let's use the relationship

                   W = ΔEm

                    W = $Em_{f}$  - Em₀

                  W = ½ k₁ x₁² - ½ k₂ (x₂ / 2)²2

                  W = ½ 511 0.1²2 - ½ 228 (0.1497 /2)²

                  W = 2,555 - 0.6387

                  W = 1.916 J