Question

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

Answer 1

$\dag\:\underline{\sf AnsWer :}$

• Here we are provided a Force (f) of 10 N and mass of the block (m) is 10 kg and Displacement (s) of the block is 5 m and we are said to calculate the work done by the block.

$:\implies \sf Work \: done = F.s \cos(\theta)$

• Force = 10 N
• Displacement = 5 m
• ⊖ = 60°

$:\implies \sf Work \: done = 10 \times 5 \cos( {60}^{ \circ} )$

$:\implies \sf Work \: done = 50 \cos( {60}^{ \circ} )$

$:\implies \sf Work \: done = 50 \times \dfrac{1}{2}$

$:\implies \sf Work \: done = \dfrac{50}{2}$

$:\implies \underline{ \boxed{ \sf Work \: done = 25 \: J}}$

Hence,the work done by the block is 25 Joule.

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

Answer 2

$\huge\bf{\pink{\underline{\underline{\mathcal{AnSwer࿐}}}}}$

Given:-

$\longrightarrow$ Force = 10 N

$\longrightarrow$ mass = 10 kg

$\longrightarrow$ Angle = 60⁰

$\longrightarrow$ displacement = 5 m

$\longrightarrow$ $\sf Cos60⁰ = \dfrac{1}{2}$

To find:-

$\longrightarrow$ Work done

Formula to be used:-

$\longrightarrow$ $\underline{\boxed{\sf work\: done = Force × displacement × CosØ}}$

Solution:-

$\implies$ $\sf Work\:done = 10 × 5 × \dfrac{1}{2}$

$\implies$ $\sf work\: done = 5 × 5$

$\implies$ $\sf work\: done = 25 Joules$

$\small\bf\pink{@itzsplendidcharm࿐}$