Question

1. A force of 10N acts on a body of mass 2kg produces a displacement of 1m with an
angle 30 the work done is​

Answer 1

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Explanation:

$given: \\ force(f) = 10 newton \\ mass(m) = 2kg \\ displacement(s) = 1m \\ angle = 30degree \\ \cos(30) = \frac{ \sqrt{3} }{2} \\ work \: done = f \times s \times \cos( \theta) \\ = 10 \times 1 \times \frac{ \sqrt{3} }{2} \\ = 5 \sqrt{3} joule \\ or \\ = 5 \times 1.73 = 8.65joule$

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Answer 2

Correct QUESTION :-

A force of 10N acts on a block of mass 10kg at an angle of 60° displaces it 5m. calculate the work done. (cos 60°=1/2).

Given:-

• Force Applied = 10N
• Mass of Block = 10kg
• Angle made = 60° = 1/2
• Displacement ( s ) = 5m

To Find:-

• Work Done.

Formulae used:-

Work Done = F × s × Cos O

Now,

→ Work Done = F × s × cos 60°

→ Work done = 10 × 5 × 1/2

→ Work done = 25J

Hence, The work done by the force is 25J.

$\begin{gathered}\bullet\:\sf Trigonometric\:Values :\\\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D{e}fined\end{tabular}}\end{gathered}$