The diagram shows a barrel of weight $ {1.0 \times 10^{3} \,N}$ on a frictionless slope inclined at $30?$ to the horizontal. The force -is parallel to the slope. What is the work done in moving the barrel a distance of $5.0 \,m$ up the slope?
- $\ce{2.5 \times 10^{3} \,J}$
- $\ce{4.3 \times 10^{3} \,J}$
- $\ce{5.0 \times 10^{3} \,J}$
- $\ce{1.0 \times 10^{3} \,J}$
The Correct Option is A
Solution and Explanation
$W = mg (h_2 - h_1)$
Here, $mg = 1.0 \times 10^3\, N$
$(h_2 - h_1) = s \, \sin \,30? = 5 \, sin\, 30?$ = 2.5 m
$\therefore \:\:\: W = 1.0 \times 10^3 \times 2.5 = 2.5 \times 10^3\, J$
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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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