Question

A 6.7 kg block is pushed into a spring that sits on a 20 degree ramp, compressing the spring 0.16 m (see the diagram below). The block is released and is measured to be moving 2.1 m/s as it passes through the relaxed position of the spring. Assume no friction is present between the block and the ramp, a) Calculate the spring constant of the spring using the energy principle, E_f = E_i + W_nc. Make sure you draw proper state diagrams for this part, b) How far (starting at the equilibrium position of the spring) along the ramp will the block go before it (momentarily) stops?

Answer 1

let

m = 6.7 kg

x = 0.16 m

theta = 20 degrees

a) Apply conservation of energy

initial mechanical energy = final mechanical energy

0.5*k*x^2 + m*g*h1 = 0.5*m*v^2 + m*g*h2

0.5*k*x^2 = 0.5*m*v^2 + m*g*(h2-h1)

0.5*k*x^2 = 0.5*m*v^2 + m*g*x*sin(theta)

k = m*v^2/x^2 + 2*m*g*sin(theta)/x

= 6.7*2.1^2/0.16^2 + 2*6.7*9.8*sin(20)/0.16

= 1435 N/m <<<<<<---------------Answer

b) Again apply conservation of energy

initial mechanical energy = final mechanical energy

0.5*k*x^2 + m*g*h1 = m*g*h2

0.5*k*x^2 = m*g*(h2-h1)

0.5*k*x^2 = m*g*d*sin(theta)

d = 0.5*k*x^2/(m*g*sin(theta))

= 0.5*1435*0.16^2/(6.7*9.8*sin(20))

= 0.818 m <<<<<<---------------Answer