A man drags a block through $10\, m$ on rough surface $(\mu = 0.5)$. A force of $\sqrt{3}$ kN acting at $30^{\circ}$ to the horizontal. The work done by applied force is
- zero
- 7.5 kJ
- 15 kJ
- 10 kJ
The Correct Option is C
Solution and Explanation
$=\left(\sqrt{3} \times 10^{3}\right) \times \cos 30^{\circ}$
$=\sqrt{3} \times 10^{3} \times \frac{\sqrt{3}}{2}$
$=\frac{3}{2} \times 10^{3} N$
Work done $= F.s$
$=\frac{3}{2} \times 10^{3} \times 10$
$=15 \times 10^{3} J$
$=15\, kJ$
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- Solutions
Concepts Used:
Work
Work is the product of the component of the force in the direction of the displacement and the magnitude of this displacement.
Work Formula:
W = Force × Distance
Where,
Work (W) is equal to the force (f) time the distance.
Work Equations:
W = F d Cos θ
Where,
W = Amount of work, F = Vector of force, D = Magnitude of displacement, and θ = Angle between the vector of force and vector of displacement.
Unit of Work:
The SI unit for the work is the joule (J), and it is defined as the work done by a force of 1 Newton in moving an object for a distance of one unit meter in the direction of the force.
Work formula is used to measure the amount of work done, force, or displacement in any maths or real-life problem. It is written as in Newton meter or Nm.
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