Energy conservation of ice problem

[paulxu11]

Homework Statement

https://moodle.telt.unsw.edu.au/pluginfile.php/2296810/question/questiontext/2691158/6/1668509/cart%20track.png
A block of ice (that we shall treat as a particle) slides with negligible friction or air resistance on the curved tpath sketched (black line). The mass of the block is m = 44 kg. Its initial speed is v0 = 1.3 m.s–1. The height h= 4.1 m. At the bottom, the path has a radius of curvature (fine circle) R = 3.4 m. At the bottom of the path, what is the force exerted on the ice by the path?

Homework Equations

K + V = constant

The Attempt at a Solution

Energy before it slides = after it slides:
mgh + 1/2 * m*v0^2 = 1/2 * m*v_final^2 (at the bottom h = 0 -> mgh=0)
44*9.8*4.1 + 1/2*44*1.3^2 = 1/2*44*v_final^2 -> v_final=9.06m/s

Circular motion:
Fc=ma=m*v_f^2/R=44*9.06^2/3.4=1062N upwards
W=mg=44*9.8=431.2N downwards
F_path=1062-431.2=630.8N upwards

Could someone suggest where I did wrong?
Thanks!

 

[paulxu11]

The pic is here

 

[vela]

Circular motion:
Fc=ma=m*v_f^2/R=44*9.06^2/3.4=1062N upwards
W=mg=44*9.8=431.2N downwards
F_path=1062-431.2=630.8N upwards

Could someone suggest where I did wrong?

If W and F_path are in opposite directions, the magnitude of the net force is 680.8 N - 431.2 N, no?

 

[haruspex]

06^2/3.4=1062N upwards
W=mg=44*9.8=431.2N downwards
F_path=1062-431.2=630.8N upwards

This is a very common error.
Centripetal force is not an applied force. It is that component of the resultant which is perpendicular to the velocity.
So in ΣF=ma, the centripetal force is the ma on the right, and the perpendicular component of each applied force appears in the sum on the left.