Question

​ a force of 10n acts on a body towards the east. the work done my the force while the body moves through 1 m in the east-north direction (mid-way between east and north) is: A 10root2J B 10/root2J ​C -10J D -10/root2J

Answer 1

Draw vector diagram of given situation you will see there is 45° between force and displacement .
w=Fdscos(Ф)
w= 10x1xcos(45°)
w=10x1x1/√2=10/√2J
option 2 is correct

Answer 2

The work done by the force is 10/root2 J .

Explanation:

Given the force acts on a body, F  = 10 N in east.

The displacement of the body, d = 1 m in east-north.

The work done is given as

$W=Fdcos\theta$

Here, $\theta$ is the angle between the force direction and movement direction.

As per question, $\theta=45^0$ (because movement direction mid-way between east and north)

so,

$W=10N\times1m(cos45^0)$

$W=\frac{10}{\sqrt{2} } J$                    ($\because cos\theta=\frac{1}{\sqrt{2} }$)

Thus, the work done by the force is 10/root2 J .

#Learn More: work done.

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