Question

A 25.0 kg is tied at the end of 10.0m cord attached to the ceiling. When displaced slightly to the side, it swing back and forth with a speed at the lowest point of 16.0 m/s

a) what is the magnitude and direction of the resultant force acting on it at this point?=640N

b)what is the tension in the cord?=885N

Answer 1

Given that

Mass of the object tied to the cord is (m) =25kg

Length of the cord (L)=(r) =10m

The speed at the lowest point is (v) =16m/s

Now the force acting at the end of the cord is one is normal force and one is weight acting downwards

Then N-mg =mv²/r

N=mv²/r+mg

=(25)(16)²/10+(25)(9.81)

=885.25N pointing upwards

The magnitue and direction of the resultant force acting on it at his point is

Fnet =N-mg =885.25 -(25)(9.81) =885.25-245.25 =640N