Question

a particle p of mass 0.4kg is moving under the action of a single force f newtons at time t seconds the velocity of p is given by v=(6t+4)i + (t*2 + 3t)j

Answer 1

The force acting on a particle of mass 0.4 kg, moving with velocity
`\( \mathbf{v} = (6t + 4)i + (t^2 + 3t)j \)`, is
`\( \mathbf{F} = 2.4i + (0.8t + 1.2)j \)` N.

The given expression for the velocity vector
`\( \mathbf{v} \) is \( \mathbf{v} = (6t + 4)i + (t^2 + 3t)j \)`. To find the force
`\( \mathbf{F} \)` acting on the particle, we can use Newton's second law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

Acceleration
`\( \mathbf{a} \)` is the derivative of velocity
`\( \mathbf{v} \)` with respect to time ( t ):

`\[ \mathbf{a} = \frac{d\mathbf{v}}{dt} \]`

Differentiating each component of
`\( \mathbf{v} \) with respect to \( t \)`gives:
`\[ \mathbf{a} = (d)/(dt)((6t + 4)i + (t^2 + 3t)j) \]`

`\[ \mathbf{a} = (6)i + (2t + 3)j \]`

Now, applying Newton's second law,
`\( \mathbf{F} = m\mathbf{a} \)`, where ( m ) is the mass of the particle (given as 0.4 kg):

`\[ \mathbf{F} = 0.4((6)i + (2t + 3)j) \]`

`\[ \mathbf{F} = 2.4i + (0.8t + 1.2)j \]`

So, the force acting on the particle is
`\( \mathbf{F} = 2.4i + (0.8t + 1.2)j \)` newtons.