2. A particle of mass 5 kg is initially at the origin. A force $$ F \mathrm{~N} $$ given by $$ F=2 s-3 $$ acts to oppose the motion, where s m is the displacement from the origin. The particle comes to rest instantaneously 5 m from the origin. Find the work done by the force in bringing the particle to rest.

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To find the work done by the force in bringing the particle to rest, we can follow these steps:

1. **Identify the known conditions:**
   - Mass of the particle, $$ m = 5 \, \text{kg} $$
   - Displacement when the particle comes to rest, $$ s = 5 \, \text{m} $$
   - Force acting on the particle, $$ F = 2s - 3 $$

2. **Calculate the force at the displacement of 5 m:**
   We will substitute $$ s = 5 $$ into the force equation to find the force at that point.

$$
   F = 2(5) - 3
   $$

3. **Calculate the work done by the force:**
   The work done by a force is given by the integral of the force over the displacement. Since the force is variable, we will calculate the work done from $$ s = 0 $$ to $$ s = 5 $$.

$$
   W = \int_{0}^{5} F \, ds
   $$

Substituting the expression for $$ F $$:

$$
   W = \int_{0}^{5} (2s - 3) \, ds
   $$

Now, let's perform the calculations step by step.

### Step 2: Calculate the force at $$ s = 5 $$

$$
F = 2(5) - 3 = 10 - 3 = 7 \, \text{N}
$$

### Step 3: Calculate the work done

Now we will calculate the work done using the integral:

$$
W = \int_{0}^{5} (2s - 3) \, ds
$$

Calculating the integral:

$$
W = \left[ s^2 - 3s \right]_{0}^{5}
$$

Now, substituting the limits:

$$
W = \left[ (5^2 - 3(5)) - (0^2 - 3(0)) \right]
$$

Calculating this gives:

$$
W = (25 - 15) - 0 = 10 \, \text{J}
$$

Thus, the work done by the force in bringing the particle to rest is $$ 10 \, \text{J} $$.
The work done by the force is 10 joules.

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