A shearing force 10 newton is applied on the top of a cuboid of height 0.1 m such that it becomes a parallelepiped. If the slant surface of the parallelepiped forms an angle of 30 degrees with the base, what is the work done by the force?

Question

Pv Aaditya · Answer

Work done = Force applied * Distance movedtan 30 = 0.1 / distance moved=> distance moved = 0.1 / (1/$$\sqrt3$$) = 0.1732 mSo, work done = 0.1732 * 10 = 1.732 joule = $$\sqrt3$$ joule

A shearing force 10 newton is applied on the top of a cuboid of height 0.1 m such that it becomes a parallelepiped. If the slant surface of the parallelepiped forms an angle of 30 degrees with the base, what is the work done by the force?

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