Question

Problem 6. A block of mass 400 grams is suspended from a vertical spring of force constant 25.0N/m. (a) How much is the stretch in the spring when the system is in equilibrium? The block is pulled a distance of 20.0 cm below its equilibrium position and released (b) What is the total energy of the system when the block is released? (Take the gravitational potential energy to be zero in the equilibrium position.) (c) What is the speed of the block when it is 12.0 cm below its equilibrium position? (d) Use an energy analysis to determine the maximum speed of the block. Does it agree with what you expect from the formula v(t) = -Aw cos(wt + po)? (e) What is the stretch or compression in the spring when the block is at the highest point of its trajectory?

Answer 1

(a)

$mg=kL$

$L=\frac {mg}{k}=\frac {0.400\times 9.8}{25}\: m = 0.1568\: m$

(b)

$U =\frac {1}{2}kx^2 + mgh$

$U =\frac {1}{2}\times 25\times 0.20^2 + 0.400\times 9.8\times 0.20$

(c)

$1.284\: J=\frac {1}{2}mv^2+0.400\times 9.8\times 0.12 + \frac {1}{2}\times 0.400\times 0.12^2$

(d)

$1.284=\frac {1}{2}mv_{max}^2$

(e)