Question

4) If the block has mass of 70 kg and its initial velocity is 3.4 m/s (at x = 3 m), determine the normal force N which the path exerts on the block right before it reaches the spring (at x = 0). Your answer must include 1 place 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.land x 3 m

Answer 1

Initial height;

y = 0.25r2 = 0.25(3)?

$\Rightarrow y =2.25$

We know;

$\Rightarrow y' =\frac{d}{dx} (0.25x^2)$

$\Rightarrow y' =0.5x$

$\Rightarrow y'' =\frac{d}{dx}(0.5x)$

$\Rightarrow y'' =0.5$

at x = 0;

Radius of curvature at the bottom;

$\rho = \frac{[1+(y')^2]^{3/2}}{|y''|}$

$\Rightarrow \rho = \frac{[1+(0)^2]^{3/2}}{|0.5|}$

$\Rightarrow \rho =2\;\;m$

By conservation of energy between initial and final point;

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

$\Rightarrow v^2 = v_o^2 + 2gy$

$\Rightarrow v^2 = 3.4^2 + 2(9.81)(2.25)$

$\Rightarrow v^2 = 55.705$

By equilibrium of forces at the bottom;

$\uparrow \sum F_n = ma_n$

$\Rightarrow N - mg = \frac{mv^2}{\rho}$

$\Rightarrow N - 70(9.81) = \frac{70(55.705)}{2}$

$\Rightarrow N =2636.4\;\;N$

...(Answer)