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A heavy-duty stapling gun uses a 0.131-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a ram spring (k = 35666 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.2 ✕ 10-2m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.3 ✕ 10-2m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

Respuesta :

Answer:

v = 15.3 m/s

Explanation:

Given:

- The mass of the metal rod m = 0.131 kg

- The spring constant of spring k = 35666 N / m

- The initial compression of the spring x_i = 3.2 ✕ 10^-2m

- The final compression of the spring x_f =  1.3 ✕ 10^-2m

Find:

Find the speed of the ram at the instant of contact.

Solution:

- The ram is initially at rest hence its initial velocity is zero. From conservation of energy principle we have and exchange of gravitational potential and elastic potential energy stored in the spring at initial and final state points:

               0.5*m*v^2 + 0.5*k*x_2^2 + m*g*h_f= 0.5*k*x_1^2 + m*g*h_o

               m*v^2 +  k*x_2^2 + 2m*g*h_f = k*x_1^2 + 2*m*g*h_o

               m*v^2 = k*x_1^2 - k*x_2^2 + 2*m*g*(h_o - h_f)

               v^2 = k/m*[x_1^2 - x_2^2] + 2*g*(h_o - h_f)

               v = sqrt {  k/m*[x_1^2 - x_2^2] + 2*g*(h_o - h_f) }

- Where, h_o: is the initial potential energy at compression state x_1.

              h_f : is the initial potential energy at compression state x_2

- Plug in the values:

              v = sqrt {  35666/.131*[0.032^2 - 0.013^2] + 2*g*(0.032 - 0.013) }

              v = sqrt {  232.7819 + 0.37278 } = sqrt {  233.1568 }

              v = 15.3 m/s

                     

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