A block with mass m = 4.5 kg is attached to two springs with spring constants kleft = 36 N/m and kright = 50 N/m. The block is pulled a distance x = 0.25 m to the left of its equilibrium position and released from rest. 1)What is the magnitude of the net force on the block (the moment it is released)?

Respuesta :

Answer:

21.5 N

Explanation:

As the block is pulled to a distance of 0.25 m to the left of its equilibrium position, the left spring is compressed by x = 0.25 m and exerting an elastic force to the right, the right spring is stretched by x =0.25 m and also exerting an elastic force to the right. Knowing their spring constant, we can calculate the total net force at that instant

[tex]F = F_{left} + F_{right}[/tex]

[tex]F = k_{left}x + k_{right}x[/tex]

[tex]F = x(k_{left} + k_{right}) = 0.25(36 + 50) = 0.25*86 = 21.5 N[/tex]

ACCESS MORE
EDU ACCESS