Question

a person pulls horizontally on blocK WITH A 5N force, at an angle of 60 with the horizontal surface. The work done in displacing block is 200 J, the displacement is

a) 80km
b)80m
c)50m
d)none of the above

Answer 1

Answer:

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80 m

SOLUTION :

GIVEN :

• Force applied to the block=5 N
• Angle on which force is applied = 60°
• Work done by the force = 200 J

TO FIND :

Displacement of the block.

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Work done by the body is equal to the product of the force applied and displacement of the body along the given angle.

Hence formula for work is —

$W = FS \cos( \alpha )$

Where —

• W is work done
• F is force applied
• S is displacement

[tex] \alpha [/tex] is angle on which force is applied

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By applying the formula and substituting the values we get

$\implies W = FS \cos( \alpha )$

$\implies 200 = 5 \times S \times \cos( 60 )$

$\implies 200 = 5 \times S \times \frac{1}{2} (since \: \cos(60) = \frac{1}{2} )$

$\implies 40 = \frac{S}{2}$

$\red{\implies 80 \: m \: = S}$

$\boxed{\boxed{Displacement \: Of \: the \: body = 80 \: m</p><p>}}$

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Answer 2

Given:-

• Applied force = F = 5 N
• As the displacement is along the horizontal, then the angle between applied force and displacement = θ = 60°
• Work done = 200 J

To find:-

• Displacement = s = ?

Answer:-

Work done is defined as the dot product of force and displacement.

W = F s cos(θ)

where,

⦾ W is work done

⦾ F is applied force

⦾ s is displacement

⦾ θ is the angle between force and displacement

Putting the values as given in the question,

W = F s cos(θ)

→ 200 = 5 × s × cos(60°)

‣ From trigonometry ratio table, we know that cos(60°) = 1/2.

→ 200 = 5 × s × (1/2)

→ [(200 × 2) / 5] = s

→ s = 80 m Ans.