When the 6kg box readies point A it has a speed nuA 10 ms D

When the 6-kg box readies point A it has a speed nu_A = 10 m/s. Determine how high the box reaches up the surface before it conics to a stop. Also, what is the resultant normal force 011 the surface at this point and the acceleration? Neglect friction and the size of the box.

Solution

apply conservation of energy

0.5 * m * v^2   = m * g *h

h = v^2 / 2g   = 100 / 2 * 9.8   = 5.1020 m

y = 5.1020 m   therefore x^0.5 + 5.1020^0.5   = 3   => x   = 0.55 m

slope = tan(theta)   = dy / dx    = -(y/x)^0.5     = -3.05

theta   = 71.832 degress from -ve x axis

acceleration   = g* sin(theta)   = 9.311 m/s^2

N = mgcos(theta)   = 6 * 9.8 * cos(71.832) = 18.334 N