A force of 50N acting on a body at an angle - with the horizontal- If 150J of work is done by displacing it through 6m- then - is-

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Standard XII

Physics

Work Energy Theorem Application

A force of ...

Question

A force of $50 N$ acting on a body at an angle $θ$ with the horizontal. If $150 J$ of work is done by displacing it through $6 m$ , then $θ$ is:

60^{o}

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30^{o}

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0^{o}

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45^{o}

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Solution

The correct option is A $60^{o}$
Horizontal component of force = $F cos θ$
$∴$ Work done = $(F cos θ) \times 6$
$150 = 50 \times cos θ \times 6$
$∴ cos θ = \frac{1}{2}$
$θ = 60^{o}$

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A force of 50N acting on a body at an angle θ with the horizontal. If 150J of work is done by displacing it through 6m, then θ is:

The correct option is A 60oHorizontal component of force =Fcosθ∴ Work done = (Fcosθ)×6 150=50×cosθ×6∴cosθ=12θ=60o

A force of $50 N$ acting on a body at an angle $θ$ with the horizontal. If $150 J$ of work is done by displacing it through $6 m$ , then $θ$ is:

The correct option is A $60^{o}$
Horizontal component of force = $F cos θ$
$∴$ Work done = $(F cos θ) \times 6$
$150 = 50 \times cos θ \times 6$
$∴ cos θ = \frac{1}{2}$
$θ = 60^{o}$