A moving perticle of mass `m` is acted upon by five forces `overset(vec)(F_(1)),overset(vec)(F_(2)),overset(vec)(F_(3)),overset(vec)(F_(4))` and `overset(vec)(F_(5))`. Forces `overset(vec)(F_(2))` and `overset(vec)(F_(3))` are conservative and their potential energy funcations are `U` and `W` respectively. Speed of the particle changes from `V_(a)` to `V_(b)` when it moves from position `a` to `b`. Choose the correct option :

Options

A. Sum of work done by `overset(vec)(F_(1)),overset(vec)(F_(4))` and `overset(vec)(F_(5)) = U_(b) - U_(a) + W_(b) - W_(a)`
B. Sum of work done by `overset(vec)(F_(1)),overset(vec)(F_(4))` and `overset(vec)(F_(5)) = U_(b) - U_(a) + W_(b) - W_(a) + (1)/(2)m(V_(b)^(2) - V_(a)^(2))`
C. Sum of work done by all five forces `= (1)/(2)m(V_(b)^(2) + V_(a)^(2))`
D. Sum of work done by `overset(vec)(F_(2))` and `overset(vec)(F_(3)) = (U_(b) + W_(b)) - (U_(a) + W_(a))`

Answer

To solve the problem, we will apply the work-energy theorem and analyze the forces acting on the particle. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Work-Energy Theorem
The work-energy theorem states that the total work done on an object is equal to the change in its kinetic energy. Mathematically, this can be expressed as:
\[ W_{\text{total}} = \Delta KE = KE_f - KE_i \]
where \( KE_f \) is the final kinetic energy and \( KE_i \) is the initial kinetic energy.

### Step 2: Write the Expression for Total Work Done
Given that the particle is acted upon by five forces \( \vec{F}_1, \vec{F}_2, \vec{F}_3, \vec{F}_4, \vec{F}_5 \), the total work done by these forces can be expressed as:
\[ W_{\text{total}} = W_1 + W_2 + W_3 + W_4 + W_5 \]

### Step 3: Calculate the Change in Kinetic Energy
The kinetic energy of the particle at position \( a \) is given by:
\[ KE_i = \frac{1}{2} m V_a^2 \]
The kinetic energy at position \( b \) is:
\[ KE_f = \frac{1}{2} m V_b^2 \]
Therefore, the change in kinetic energy is:
\[ \Delta KE = \frac{1}{2} m V_b^2 - \frac{1}{2} m V_a^2 \]

### Step 4: Relate Work Done to Change in Kinetic Energy
By substituting the expression for change in kinetic energy into the work-energy theorem, we have:
\[ W_{\text{total}} = \frac{1}{2} m V_b^2 - \frac{1}{2} m V_a^2 \]

### Step 5: Analyze the Work Done by Conservative Forces
For the conservative forces \( \vec{F}_2 \) and \( \vec{F}_3 \), the work done can be expressed in terms of potential energy:
\[ W_2 = -\Delta U = - (U_b - U_a) = U_a - U_b \]
\[ W_3 = -\Delta W = - (W_b - W_a) = W_a - W_b \]
Thus, the total work done by the conservative forces is:
\[ W_2 + W_3 = (U_a - U_b) + (W_a - W_b) \]

### Step 6: Calculate Work Done by Non-Conservative Forces
The work done by the non-conservative forces \( \vec{F}_1, \vec{F}_4, \vec{F}_5 \) can be expressed as:
\[ W_1 + W_4 + W_5 = W_{\text{total}} - (W_2 + W_3) \]

### Step 7: Substitute and Rearrange
Substituting the expressions for \( W_{\text{total}} \) and \( W_2 + W_3 \):
\[ W_1 + W_4 + W_5 = \left(\frac{1}{2} m V_b^2 - \frac{1}{2} m V_a^2\right) - \left((U_a - U_b) + (W_a - W_b)\right) \]
This simplifies to:
\[ W_1 + W_4 + W_5 = \frac{1}{2} m V_b^2 - \frac{1}{2} m V_a^2 + U_b - U_a + W_b - W_a \]

### Conclusion
Thus, we have derived the expressions for the work done by the forces acting on the particle and can choose the correct options based on the analysis.