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A block of mass 5 kg slides down a rough inclined surface. The angle of inclination is `45^(@)`. The coefficient of sliding friction is 0.20. When the block slides 10 cm, the work done on the block by force of friction isA. `-(1)/(sqrt2)J`B. 1JC. `-sqrt2J`D. `-1J` |
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Answer» Correct Answer - A (a) Work done by frictional force, `W_(f)=fs cos 180^(@)` `=(mu mg cos theta)(-1)(s)` `=-0.2xx5xx10xx(1)/(sqrt2)xx0.1=-(1)/(sqrt2)J` |
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