Question

a force 10N acts on a block ob mass 10kg at an angle of 60 displaces it 5meter calculate work (cas 60=1/2​

Answer 1

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Work done in fruit case =F.S.cosθ

=10×2×cos60

=10J

Work done in second case is same as 10J=FScosθ

⇒F×2×cos30=10

F= 10/√3N

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Answer 2

Explanation:

Given :-

A force 10 N acts on a block af mass 10 kg at an angle of 60° displaces it 5 metre. (cos 60° = ½)

To Find :-

What is the work.

Formula Used :-

{\red{\boxed{\large{\bold{Work\: Done\: =\: Fs\: Cos\theta}}}}}

WorkDone=FsCosθ

where,

F = Force

s = Displacement

\thetaθ = Angle between applied force and displacement.

Solution :-

Given :

Force = 10 N

Displacement = 5 m

According to the question by using the formula we get,

⇒ \sf Work\: Done =\: Fs\: Cos\thetaWorkDone=FsCosθ

( As we know that, Cos\theta =\: \dfrac{1}{2}θ=

2

1

)

⇒ \sf Work\: Done =\: {\cancel{10}} \times 5 \times \dfrac{1}{\cancel{2}}WorkDone=

10

×5×

2

1

⇒ \sf Work\: Done =\: 5 \times 5WorkDone=5×5

➠ \sf\bold{\purple{Work\: Done =\: 25\: J}}WorkDone=25J

\therefore∴ The work done is 25 J .