Question

A boy pulls a toy with a force of 50N through a string which makes an angle of 30 with the horizontal so as to move the toy by a distance of 1m horizontal. Of the string were inclined making an angle of 45 with the horizontal, how much pull would be apply along the string in order to move it the same distance of 1m?

Answer 1

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

★Given

F = 50N

θ = 30

s = 1m

Work done = F × S × cosθ

= 50N × 1N × cos30

= 50 × 1 × 0.866j

= 43.3j

$\text{Second\;situation\;when\;force\;applied\;at\;angle\; of\;45}$

Work done = F × S × cosθ

43.3j = F × S × cos45

43.3j = F × 1m × cos45

$\tt{\rightarrow F=\dfrac{43.3j}{1m\times cos45}}$

$\tt{\rightarrow\dfrac{43.3mm}{0.707m}}$

= 61.2N

$\boxed{\begin{minipage}{11 cm} Additional Information \\ \\ \ Force=Mass\times Acceleration \\ \\ Kinetic Energy = \dfrac{1}{2}\times Mass\times (Velocity)^{2} \\ \\ Potential\; Energy=Mass\times Acceleration\;due\;to\;gravity\times Height \\ \\ Power = \dfrac{Total\;Work\;done}{Time}=\dfrac{Energy}{Time} \end{minipage}}$

Answer 2

》 A boy pulls a toy with force of 50N through a string which makes an angle of 30° with the horizontal. So, as to move the toy by a distance of 1 m m horizontal.

Here..

• Force (F) = 50 N

• Angle (Ø) = 30°

• Distance (s) = 1 m

______________ [ GIVEN ]

• We have to find the work done.

________________________________

We know that..

Work done = F × s cosØ

Put the known values in above formula

=> Work done = 50 × 1 × $\dfrac{ \sqrt{3} }{2}$

=> $25\sqrt{3}$ J

________________________________

》 Now the string were inclined making an angle of 45 with the horizontal with a distance of 1 m.

Here..

• Work done = $25\sqrt{3}$ J

• Distance (s) = 1 m

• Angle (Ø) = 45°

__________ [ GIVEN ]

• We have to find the force (F).

________________________________

Work done = F × s cosØ

=> $25\sqrt{3}$ = F × 1 × $\dfrac{ \sqrt{2} }{2}$

=> $25\sqrt{3}$ = F × $\dfrac{ \sqrt{2} }{2}$

=> F = $25\sqrt{3}$ × $\dfrac{ 2 }{\sqrt{2}}$

=> F = $\dfrac{50 \sqrt{3} }{ \sqrt{2} }$

After rationalization we get

=> F = $\dfrac{50 \sqrt{6} }{ 2}$

=> F = $25\sqrt{6}$ N

______________________________

Force = $25\sqrt{6}$ N or 61 N (approx.)

_____________ [ ANSWER ]

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