A uniform force of $(3\hat{i}+\hat{j}) $newton acts on a particle of mass 2 kg. Hence the particle is displaced from position $(2\hat{i}+\hat{k})$ metre to position $(4\hat{i}+3\hat{j}-\hat{k})$ metre. The work done by the force on the particle is

Question

In a Vernier caliper, $N+1$ divisions of vernier scale coincide with $N$ divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:

Answer 1

To find the work done by the force on the particle, we can use the formula for work done by a constant force, which is defined as the dot product of the force vector and the displacement vector. The formula is:

W = \mathbf{F} \cdot \mathbf{s}

Here, \mathbf{F} is the force vector and \mathbf{s} is the displacement vector.

Given:

Force vector, \mathbf{F} = 3\hat{i} + \hat{j}
Initial position, \mathbf{r}_1 = 2\hat{i} + \hat{k}
Final position, \mathbf{r}_2 = 4\hat{i} + 3\hat{j} - \hat{k}

The displacement vector \mathbf{s} is given by:

\mathbf{s} = \mathbf{r}_2 - \mathbf{r}_1

Substituting the values, we get:

\mathbf{s} = (4\hat{i} + 3\hat{j} - \hat{k}) - (2\hat{i} + \hat{k})

Simplifying,

\mathbf{s} = (4 - 2)\hat{i} + (3 - 0)\hat{j} + (-1 - 1)\hat{k}

\mathbf{s} = 2\hat{i} + 3\hat{j} - 2\hat{k}

Now, calculate the work done:

W = \mathbf{F} \cdot \mathbf{s} = (3\hat{i} + \hat{j}) \cdot (2\hat{i} + 3\hat{j} - 2\hat{k})

The dot product is calculated as follows:

W = (3 \times 2) + (1 \times 3) + (0 \times -2)

W = 6 + 3 + 0

W = 9

Thus, the work done by the force on the particle is 9\hat{J}.

Answer 2