A block of mass $$5$$ kg resting on a horizontal surface is connected by a cord, passing over a light frictionless pulley to a hanging block of mass $$5$$ kg. The coefficient of kinetic friction between the block and the surface is $$0.5$$. Tension in the cord is (Take$$, g $$=$$ $$9.8$$ $$ms-2)

Question

Infinity Learn · Accepted Answer

⇒ $$T-f$$ $$=$$ $$5a$$, $$T-$$μmg $$=$$ $$5aT-0.5$$×$$5$$×$$9.8$$ $$=$$ $$5a$$          ......(i)⇒   $$5g$$ $$-$$ T $$=$$ $$5a$$        ......(ii)From$$ Eqs. $$(i) and (ii), we $$get5$$×$$9.8-0.5$$×$$5$$×$$9.8=10a$$⇒  $$0.5$$×$$5$$×$$9.8=10a$$⇒  $$a=2.45$$ $$m/s2From$$ Eq (ii), we $$get5$$×$$9.8-5$$×$$2.45=T$$⇒  $$T=36.75$$ N

A block of mass 5 kg resting on a horizontal surface is connected by a cord, passing over a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the block and the surface is 0.5. Tension in the cord is (Take, g = 9.8 ms^-2)

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A block of mass 5 kg resting on a horizontal surface is connected by a cord, passing over a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the block and the surface is 0.5. Tension in the cord is (Take, g = 9.8 ms-2)

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A block of mass 5 kg resting on a horizontal surface is connected by a cord, passing over a light frictionless pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the block and the surface is 0.5. Tension in the cord is (Take, g = 9.8 ms^-2)