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<span>the spring constant is 200kN/m.
</span>
The work done on the spring by the the free falling safe is equal to the kinetic energy of the spring:
W1 =
k
The work done on the safe by gravity is the potential energy of the spring:
W2 = (mass of safe) x g x (height)
= 1000kg x 9.8 m/s^2 x (2 + 0.54)m
=1000kg x 9.8 m/s^2 x (2.54)m
= 24,892 N-m
W2 = W1 =24892 N-m
⇒k
= 24892 N-m
Therefore, k =
= 199136 kN/m.
The spring constant is 199,136 N/m ≈ 200 kN/m
</span>
The work done on the spring by the the free falling safe is equal to the kinetic energy of the spring:
W1 =
The work done on the safe by gravity is the potential energy of the spring:
W2 = (mass of safe) x g x (height)
= 1000kg x 9.8 m/s^2 x (2 + 0.54)m
=1000kg x 9.8 m/s^2 x (2.54)m
= 24,892 N-m
W2 = W1 =24892 N-m
⇒
Therefore, k =
The spring constant is 199,136 N/m ≈ 200 kN/m
Answer:
|Fr| = 50*√2 [N]
Explanation: (See Annex )
En el diagrama de cuerpo libre ( mostrado en el anexo) se ve que, al ser las dos fuerzas iguales (50 N) el paralelogramo formado para encontrar la resultante de las fuerzas (Fr) es un cuadrado, y la diagonal de ese cuadrado ( que es al mismo tiempo la hipotenusa del triangulo recto OPA ) es igual a:
Hipotenusa = |Fr| = √ (50)² + (50)²
|Fr| = √2* (50)²
|Fr| = 50*√2 [N]
