a block of mass 2.0 kg slides on arough surface at.t=0.its speed is 2.0km/s it stopa after convening a distance of 20 cm because of the friction
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Solution
initial Kinetic energy , K1=12mv21=12×2×2=2 JK1=12mv12=12×2×2=2 J
Final Kinetic energy , K2=0K2=0
Using work energy theorem we get
Work done by friction= Change in kinetic energy = K2−K1=−2 J
Explanation:
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