A smooth track in the form of a quarter-circle of radius 6 m lies in the vertical plane. A ring of weight 4(N) moves from `P_(1)` and `P_(2)` under the action of forces `F_(1) , F_2` and `F_3`. is always towards `P_(2)` under the action of forces `F_(1),F_(2)` and `F_(3)` Force `F_(1)` is always towards `P_(2)` and is always (20) N in maghitude, force `F_(2)` always acts hotizontally and is always (30 N) in magnitude, force `F_(3)` always acts tangentially to the track and is of magnitude `(15-10 s) N`, where s is in metre. If the particle has speed `4 m//s` at `P_(1)`,what will its speed be at `P_(2)`?

Explanation

To solve the problem step by step, we will analyze the forces acting on the ring and calculate the work done by each force as the ring moves from point P1 to P2. We will then apply the work-energy principle to find the speed of the ring at P2.

### Step 1: Identify the Forces Acting on the Ring
The forces acting on the ring are:
- \( F_1 = 20 \, \text{N} \) (towards P2)
- \( F_2 = 30 \, \text{N} \) (horizontal)
- \( F_3 = 15 - 10s \, \text{N} \) (tangential, where \( s \) is the distance in meters)

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