Using the law of conservation of energy, determine the velocity and the normal reaction of the triangular crate when it is at the centre of the arc with a radius of curvature of 5 m, if the mass of the crate is 5 kg. The crate is sliding on the smooth curved ramp and it starts sliding the curved ramp with a velocity of 20 m/s (at A).

1) 18.8 m/s & 96 N
2) 19.5 m/s & 76 N
3) 14.5 m/s & 65 N
4) 16.5 m/s & 50 N

Assuming B as the centre of the arc and applying the conservation of energy principle we have:
TA +VA = TB + VB
1/2 (5) (20)2 = 5 (10) + 1/2 (5) (VB)2
VB= 19.5 m/s

So the velocity at B is VB= 19.5 m/s

Summing up all the forces acting at point B we have :
S Fn = man
NB = 2 (19.5)2/10
NB = 76 N

So the velocity of the crate at B is VB= 19.5 m/s and
the normal force acting at B is NB = 76 N