A body of mass 5 kg under the action of constant force ` vec(F)=F_(x)hati+F_(y)hatj` has velocity at t=0 s as `vecv=(6hati-2hatj)` m/s and at t=10s as `vecv=+6hatj` m/s. The force `vecF` is:

Options

A. `(-3hati+4hatj)N`
B. `(-(3)/(5)hati+(4)/(5)hatj)N`
C. `(3hati-4hatj)N`
D. `((3)/(5)hati-(4)/(5)hatj)N`

Explanation

To solve the problem, we need to find the force vector \(\vec{F}\) acting on a body of mass \(5 \, \text{kg}\) given its initial and final velocities over a time interval. Here are the steps to find the force vector:

### Step 1: Identify the given information
- Mass of the body, \(m = 5 \, \text{kg}\)
- Initial velocity at \(t = 0 \, \text{s}\), \(\vec{v}_0 = 6 \hat{i} - 2 \hat{j} \, \text{m/s}\)
- Final velocity at \(t = 10 \, \text{s}\), \(\vec{v}_f = 0 \hat{i} + 6 \hat{j} \, \text{m/s}\)

### Step 2: Calculate the change in velocity
The change in velocity \(\Delta \vec{v}\) can be calculated as:
\[ \Delta \vec{v} = \vec{v}_f - \vec{v}_0 \]
Substituting the values:
\[ \Delta \vec{v} = (0 \hat{i} + 6 \hat{j}) - (6 \hat{i} - 2 \hat{j}) = -6 \hat{i} + 8 \hat{j} \, \text{m/s} \]

### Step 3: Calculate the acceleration
Acceleration \(\vec{a}\) is defined as the change in velocity over time:
\[ \vec{a} = \frac{\Delta \vec{v}}{\Delta t} \]
Here, \(\Delta t = 10 \, \text{s}\):
\[ \vec{a} = \frac{-6 \hat{i} + 8 \hat{j}}{10} = -0.6 \hat{i} + 0.8 \hat{j} \, \text{m/s}^2 \]

### Step 4: Calculate the force vector
Using Newton's second law, the force \(\vec{F}\) can be calculated as:
\[ \vec{F} = m \vec{a} \]
Substituting the values:
\[ \vec{F} = 5 \, \text{kg} \cdot (-0.6 \hat{i} + 0.8 \hat{j}) = -3 \hat{i} + 4 \hat{j} \, \text{N} \]

### Final Answer
Thus, the force vector is:
\[ \vec{F} = -3 \hat{i} + 4 \hat{j} \, \text{N} \]

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