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A force of 45 N acts on a body and moves it a distance of 3m on a horizontal surface. Calculate the work done if the direction of force is at an angle of 60° to the horizontal surface. Derive the relation between kWh and joule. |
Answer» <html><body><p>tion:Force acting on the body = 10 kgf = 10 x 10 <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> = 100 NDisplacement, S=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.5 mWork done= force x displacement in the direction of force(i)W =F x SW = 100 x 0.5= 50 <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>(ii)<a href="https://interviewquestions.tuteehub.com/tag/work-20377" style="font-weight:bold;" target="_blank" title="Click to know more about WORK">WORK</a> = force x displacement in the direction of forceW = F x S cosW = 100 x 0.5 cos60oW= 100 x0.5 x 0.5(cos60o=0.5)W=25 J(iii)<a href="https://interviewquestions.tuteehub.com/tag/normal-1123860" style="font-weight:bold;" target="_blank" title="Click to know more about NORMAL">NORMAL</a> to the force:Work = force x displacement in the direction of forceW = F x S cosW = 100 x 0.5 cos90oW= 100 x 0.5 x 0 =0 J(cos90o =0)</p></body></html> | |