Question

a particle has mass 3kg and moves under a force of 4i+8j+10k N .if particle starts from rest and was at the origin ,what are its new co-ordinates after 3s.

Answer 1

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Answer 2

Answer:

The new coordinate of a particle after 3 seconds is (6, 12, 15)

Explanation:

Given that,

Force $F = (4i+8j+10k) N$

Mass m = 3 kg

Time t = 3 s

The force is

$F =(4i+8j+10k) N$

The force along x-axis

$F_{x} = 4 N$

The acceleration is

$a_{x} = \dfrac{F_{x}}{m}$

$a_{x} = \dfrac{4\ N}{3\ kg} = 1.33m/s^{2}$

Now, using equation of motion

$s_{x}=u_{x}t+\dfrac{1}{2}a_{x}t^{2}$

$s_{x}\text{ is x coordinate of the particle.}$

$s_{x} = 0+\dfrac{1}{2}\times 1.33\times 9$

$s_{x} = 5.98\approx 6\ m$

The force along y-axis

$F_{y} = 8 N$

The acceleration is

$a_{y} = \dfrac{F_{y}}{m}$

$a_{y} = \dfrac{8\ N}{3\ kg} = 2.67m/s^{2}$

Now, using equation of motion

$s_{y}=u_{y}t+\dfrac{1}{2}a_{y}t^{2}$

$s_{y}\text{ is y coordinate of the particle.}$

$s_{y} = 0+\dfrac{1}{2}\times 2.67\times 9$

$s_{y} = 12.02\approx 12\ m$

The force along z-axis

$F_{y} = 10 N$

The acceleration is

$a_{z} = \dfrac{F_{z}}{m}$

$a_{z} = \dfrac{10\ N}{3\ kg} = 3.33m/s^{2}$

Now, using equation of motion

$s_{z}=u_{z}t+\dfrac{1}{2}a_{z}t^{2}$

$s_{z}\text{ is z coordinate of the particle.}$

$s_{z} = 0+\dfrac{1}{2}\times 3.33\times 9$

$s_{z} = 14.98\approx 15\ m$

Hence, The new coordinate of a particle after 3 seconds is (6, 12, 15)