Question

how u can solve it ✌️✌️​

Answer 1

Question:

A force F is applied on a block at an angle θ. The block is having a displacement x. If work done by the force is Fx/2, then the angle θ is ____

Answer:

Angle θ = 60°

Explanation:

Given:

• Force = F
• Displacement = x
• Work done = Fx/2

To find:

Value of angle θ = ?

★ Formula used:

To solve this question, we use the formula of Work done.

$\red{\boldsymbol{Work \: \: done = (Force) \times (Displacement) \times (cos \theta)}}$

★ Solution:

Applying the above formula,

$\implies \sf{ \dfrac{Fx}{2} = (F) \times (x) \times (cos \theta) }$

Cancelling F and x on both sides,

$\longrightarrow \: \: \sf{ \dfrac{1}{2} = cos \theta }$

Taking Cos inverse on both sides,

$\longrightarrow \: \: \sf{\theta = cos^{ - 1} \bigg( \dfrac{1}{2} \bigg) }$

$\longrightarrow \: \: \sf{ \theta = 60^{ \circ} }$

$\therefore \: \boxed{\underline{\bf{Angle \: \: \theta} = {60}^{ \circ} }}$

Answer 2

Answer:

Answer:

Angle θ = 60°

\begin{gathered}\\\end{gathered}

Explanation:

Given:

Force = F

Displacement = x

Work done = Fx/2

\begin{gathered}\\\end{gathered}

To find:

Value of angle θ = ?

\begin{gathered}\\\end{gathered}

★ Formula used:

To solve this question, we use the formula of Work done.

\begin{gathered} \red{\boldsymbol{Work \: \: done = (Force) \times (Displacement) \times (cos \theta)}} \\ \\ \end{gathered}

Workdone=(Force)×(Displacement)×(cosθ)

★ Solution:

Applying the above formula,

\begin{gathered} \implies \sf{ \dfrac{Fx}{2} = (F) \times (x) \times (cos \theta) } \\ \\ \end{gathered}

⟹

2

Fx

=(F)×(x)×(cosθ)

Cancelling F and x on both sides,

\begin{gathered} \longrightarrow \: \: \sf{ \dfrac{1}{2} = cos \theta } \\ \\ \end{gathered}

⟶

2

1

=cosθ

Taking Cos inverse on both sides,

\begin{gathered}\longrightarrow \: \: \sf{\theta = cos^{ - 1} \bigg( \dfrac{1}{2} \bigg) } \\ \\ \end{gathered}

⟶θ=cos

−1

(

2

1

)

\begin{gathered} \longrightarrow \: \: \sf{ \theta = 60^{ \circ} } \\ \\ \end{gathered}

⟶θ=60

∘

\begin{gathered} \therefore \: \boxed{\underline{\bf{Angle \: \: \theta} = {60}^{ \circ} }} \\ \\ \end{gathered}

∴

Angleθ=60

∘