Question

3. For this series of 5 questions, a block is released from the position shown with a certain initial velocity. It then slides down the smooth (friction-less) path for which the function is given in the image. You will be asked to solve for different things. Please pay attention: the numbers may change from problem to problem since they are randomized. 2) As the block slides down, its potential energy converts into kinetic energy. If its initial velocity is 1.5 m/s (at x = 3 m), determine its velocity, v, at x = 0.0 m. Your answer must include 2 places after the decimal point and the proper unit. Take g = 9.81 m/s2 у v = 1 m/s y = 0.25x2 k = 5 kN/m www х 3 m

Answer 1

Initial height at x = 3 m is;

y = 0.25r2 = 0.25(3)?

$\Rightarrow y =2.25\;\;m$

By conservation of energy;

$\frac{1}{2}mv_o^2 +mgy = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{v_o^2+2gy}$

#v= V1.52 + 2(9.81) (2.25)

$\Rightarrow v =6.81\;\;m/s$

...(Answer)