Question

Body of mass 2 kg is moving towards east with a uniform speed of 4 metre per second a force of 4 newton is applied to it towards the north what is the magnitude of displacement after 2 second

Answer 1

Answer:

Explanation:

u = 2i

a= F/m=j

S= (2i)2 +1/2(j)4

=4i +2j

Magnitude of S = square root(20) m

= 4.5 m

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Answer 2

The magnitude of displacement after 2 second is 8.94 meters.

Explanation:

Given that,

Mass of body, m = 2 kg

Velocity of body towards east, $v_x=4\ m/s$

Force applied towards North is, $F_y=4\ N$

Acceleration towards y direction is :

$F_y=ma_y\\\\a_y=\dfrac{F_y}{m}\\\\a_y=\dfrac{4}{2}\\\\a_y=2\ m/s^2$

Displacement after 2 seconds in y direction is :

$y=\dfrac{1}{2}a_yt^2\\\\y=\dfrac{1}{2}\times 2\times 2^2=4\ m$

Displacement after 2 seconds in x direction :

$x=v_xt\\\\x=4\times 2\\\\x=8\ m/s$

Net displacement after 2 seconds is :

$d=\sqrt{x^2+y^2} \\\\d=\sqrt{8^2+4^2} \\\\d=8.94\ m$

So, the magnitude of displacement after 2 second is 8.94 meters.

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Displacement