# Mechanics question help ?

## A force of 6i+j N acts on a particle of mass 2 kg. The initial velocity of the particle is 2i−5j ms-1. The velocity of the particle after 4 seconds is _i + ____j ms-1.

Mar 31, 2018

See below.

#### Explanation:

$m \left(\ddot{x} , \ddot{y}\right) = \left(0 , - m g\right) + \left(6 , 1\right)$

integrating once

$m \left(\dot{x} , \dot{y}\right) = m \left({v}_{x} , {v}_{y}\right) = \left(0 , - m g\right) t + \left(6 , 1\right) t + m \left({v}_{0 x} , {v}_{0 y}\right)$

but $\left({v}_{0 x} , {v}_{0 y}\right) = \left(2 , - 5\right)$ then

$\left({v}_{x} , {v}_{y}\right) = \left(0 , - g\right) t + \frac{1}{m} \left(6 , 1\right) t + \left(2 , - 5\right)$ and after $4$ seconds

$\left({v}_{x} , {v}_{y}\right) = 4 \times \left(0 , - g\right) + 4 \times \frac{1}{m} \left(6 , 1\right) + \left(2 , - 5\right)$

Apr 2, 2018

#### Answer:

$\left(14 \hat{i} - 3 \hat{j}\right) \setminus \text{m/s}$

#### Explanation:

Force on particle is

$\text{F" = (6hati + hatj)\ "N}$

Acceleration of particle is

${\text{a" = "F"/"m" = ((6hati + hatj)\ "N")/"2 kg" = (3hati + 0.5hatj)\ "m/s}}^{2}$

Velocity of particle after $\text{4 s}$ is

$\text{v = u + at}$

"v" = [(2hati - 5hatj)\ "m/s"] + [(3hati + 0.5hatj)\ "m/s"^cancel(2) × 4 cancel"s"]

$\text{v" = (2hati - 5hatj)\ "m/s" + (12hati + 2hatj)\ "m/s}$

$\text{v" = (14hati - 3hatj)\ "m/s}$