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Solution to your asked query is provided in tabular form:
θ Cosθ W = F.s cos θ Conclusion
0o  Cos 0= 1 W = F.s  when θ =0then value of cos 0o is 1therefore workdone = force×displacement ×cos 0oW =F.s.1 =F.sthis implies that workdone is positive.
90o  Cos 90o = 0 W = 0 when θ = 90o then the value of cosθ = 0then workdone is also equals to 0 becauseWorkdone( W) = Force(f) ×displacement(s) ×cosθas cos 90o =0 therefore W = 0
180o  Cos 180o = - 1 W = - F.s when θ =180othen value ofcos 1800 =-1therefore W =F×s×cosθW =F×s×cos 1800 W=F×s×(-1)W=-F.sthis implies that workdone is negative

Hope this information clears your doubt about the topic.

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