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
