Question

A body moves a distance of 10 metre along a straight line under the action of 5 newton force if the work done is 25 then the angle between the force and the direction of motion of the body is

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Distance (s) = 10 meters.
• Force acting on it (F) = 5 N.
• Angle b/w Force and Displacement (θ) = ?.
• Work done by the force (W) = 25 J.

$\huge{\bold{\underline{Explanation:-}}}$

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As we Know Work done formula,

$\large{\boxed{\tt W = F.S \cos \theta}}$

(Here Dot product of vectors is involved)

Substituting the values,

$\large{\tt 25 = 10 \times 5 \times \cos \theta}$

Now,

$\large{\tt 25 = 50 \times \cos \theta}$

$\large{\tt \cos \theta = \dfrac{25}{50}}$

$\large{\tt \cos \theta = \dfrac{\cancel{25}}{\cancel{50}}}$

$\large{\tt \cos \theta = \dfrac{1}{2}}$

Therefore,

$\large{\tt \theta = \cos^{-1} \dfrac{1}{2}}$

It becomes,

$\huge{\boxed{\boxed{\tt \theta = 60 }}}$

So, the Angle between Force and displacement is 60°

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Important Formulas:-

• ${\tt K.E = \dfrac{1}{2}mv^2}$
• ${\tt P.E = mgh}$
• ${\tt U = \dfrac{1}{2}Kx^2}$ (For Spring Potential energy.)
• $\large{\tt W = \displaystyle\int F.ds}$ (For a variable Force.)
• ${\tt W = K.E_f - K.E_i}$ (By Work - energy theorem. )

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Answer 2

$\Huge{\star}{\underline{\boxed{\red{\sf{Answer :}}}}}{\star}$

Given :-

Distance (s) = 10 m

Force (F) = 5 N

Work Done (W) = 25 J

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To Find :-

Angle between force and distance

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Solution :-

We have formula :-

$\Huge{\boxed{\boxed{\sf{W \: = \: F.S \: Cos \theta}}}}$

______________[Put Values]

$\Large \implies {\sf{25 \: = \: 10.5 \: Cos \theta}}$

$\Large \implies {\sf{25 \: = \: 50 \: Cos \theta}}$

$\Large \implies {\sf{Cos \theta \: = \: \frac{\cancel{25}}{\cancel{50}}}}$

$\Large \implies {\sf{Cos \theta \: = \: \frac{1}{2}}}$

As we know that 1/2 is Cos 60°

So,

$\Large \implies {\sf{\cancel{Cos} \theta \: = \: \cancel{Cos} 60^{\circ}}}$

$\Huge{\boxed{\boxed{\sf{ \theta \: = \: 60^{\circ}}}}}$