Question

My friend has applied a force of 1250N on a wooden block then the angle an be form 30 degree to the horizontal direction and 2050J work can be done then what is the displacement.​

Answer 1

Answer:

1.88 metres is the required displacement.

Explanation:

According to the Question

It is given that ,

• Force ,F = 1250N
• Work Done ,W = 2050J
• Angle ,∅ = 30°

we need to calculate the displacement of the wooden block.As we know that work done is calculated by the product of Force and displacement.

• W = FsCos∅

where,

W denote work done

F denote Force applied

∅ denote Angle

Substitute the value we get

➻ 2050 = 1250×s Cos30°

➻ 2050 = 1250 × s × √3/2

➻ 2050 = 1250 × √3/2 × s

➻ 2050/1250 = √3/2 × s

➻ 1.64 = 0.87 ×s

➻ 1.64/0.87 = s

➻ s = 1.88m

• Hence, the displacement of the wooden block will be 1.88 metres (approx).

Answer 2

$\large \sf {\underbrace{\underline{Elucidation:}}}$

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$\rm \red {\underline{\underline{provided\: that:}}}$

➻Force applied(F)=1250N

➻Angle θ=30°

➻Work done(W)=2050J

━═━═━═━═━═━═━═━═━═━═━═━═━═━

$\rm \blue {\underline{\underline{To\: be\: determined:}}}$

➻Displacement (s)=?

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$\rm \pink {\underline{\underline{We\: know:}}}$

$\implies \tt {\fbox{W=Fscosθ}}$

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➻Supplanting the values given,

$\colon \mapsto \tt {W=Fscosθ}$

$\colon \mapsto \tt {2050J=1250N\times s\times cos(30°)}$

$\large \colon \mapsto \tt {2050J=1250N\times s\times {\frac{\sqrt{3}}{2}}}$

$\small \tt {[∵ Cos 30°=\frac{\sqrt{3}}{2}]}$

$\large \colon \mapsto \tt {s=\frac{2050J\times 2}{1250N\times{\sqrt{3}}}}$

$\large \colon \mapsto \tt {s=\frac{2050J\times 2}{1250N\times1.732}}$

$\small \tt {[∵ Value\: of{\sqrt{3}}=1.732]}$

$\large \colon \mapsto \tt {s=\frac{4100J}{2165N}}$

$\large \colon \mapsto \tt {s=1.89J/N^{-1}}$

$\colon \implies \tt \green {\fbox{s=1.89m}}$

━═━═━═━═━═━═━═━═━═━═━═━═━═━

$\rm \purple {\underline{\underline{Thereupon,}}}$

➻Displacement (s)=1.89m.

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