Question

A small block of mass 1 kg is moving with constant speed of 10 m/s on a typical path in a xy vertical plane whose equation is $y = \frac{x^3}{30} - \frac{x^2}{10}$. Here, y-axis is along the vertical upward direction and x-axis along horizontal. Normal reaction (in newton) offered by ground at x = 2 m is 10K. Find K. Take g = 10 m/s²

Answer 1

0

Answer 2

The path's curvature at $x=2$ m is calculated using derivatives. $y'(2)=0$ implies a horizontal tangent, and $y''(2)=1/5>0$ implies upward concavity. For a horizontal tangent, upward concavity means the center of curvature is below the point, so centripetal acceleration ($a_c=v^2/R = 20$ m/s²) is downwards. Forces are weight $mg=10$ N downwards and normal force $N$ upwards. The equation of motion is $N - mg = -m a_c$. Substituting values gives $N = mg - m a_c = 10 - 20 = -10$ N. A negative normal force signifies that the block would lift off the ground. Therefore, the actual normal reaction offered by the ground is 0 N. Given $N=10K$, we get $10K=0$, so $K=0$.