Question

Ley de Hooke

Answer 1

Answer:

1)  F = 58.45 N ,  2) m = 4,490 kg ,  3)  x = -0.10 m , 4)   k = 1750 N / m

5)  k = 19.6 N / m

Explanation:

This problem is formed by small exercises regarding Hooke's law and Young's modulus

Hooke's law

1) The elastic force is

         F = k x

         F = 835 0.07

         F = 58.45 N

2) from the condition of static equilibrium the spring force is equal to the hanging weight

          W = F

          m g = k x

          m = k x / g

          m = 1100 0.04 / 9.8

          m = 4,490 kg

3) F = -k x

          x = - F / k

         x = - 10/100

         x = -0.10 m

4) F = k x

         k = F / x

         k = 700 / 0.40

         k = 1750 N / m

5) k = 19.6 / 1

         k = 19.6 N / m

Young's modulus

the expression for the unit elongation a one dimension is

             F / A = Y Dx / L

1) they ask us for the Young module

let's look for the area the wire

          A = pi r2 = pi d2 / 4

          A = pi 0.0022 2/4

          A = 3.80 10-6 m²

         Y = (F / A) / (Dx / L)

we substitute

         Y = (390 / 3.90 10-6) / (0.10 / 120)

         Y = 1 10-4 / 8.33 10-4

         Y = 0.12 N / m²

2) F / A = Y Dx / L

the young modulus of steel is 20 10 10 N / m²

           F / A = 20 1010 0.003 / 2

           F / A = 3 108 N / m²

3) marine force occurs when elongation is equal to initial length

               F / A = Y 1

               F = Y A

               F = 20 1010 3 10-4

               F = 6 10 7 N

4) Dx = L F / (A Y)

           W = F = m g

    Dx = 1.2 (400 9.8) / 20 1010 3 10-4)

            Dx = 7.84 10-5 m