A lawn mower is pushed a horizontal distance of 20 m by a force of 200 N directed at an angle of 300 with the ground. What is the work of this force?
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Answered by
23
Given :
Displacement (S) = 20 m
Force (F) = 200 N
Angle (theta) = 30°
Work done = F.S.cos(theta)
= 200 × 20 × cos30°
= 4000 × (√3/2)
= 2000√3 J
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Displacement (S) = 20 m
Force (F) = 200 N
Angle (theta) = 30°
Work done = F.S.cos(theta)
= 200 × 20 × cos30°
= 4000 × (√3/2)
= 2000√3 J
________________________
Dhruv00:
angle is 300° not 30°,so edit it
sorry but it is 30 only
Yeah , I just guessed it : )
its OK
the given answer is correct or what
Mine answer is absolutely correct then !
Answered by
5
AS WE KNOW THAT
WORK DONE=FORCE × DISPLACEMENT × COS@
WORK=200 ×20× cos 30
Work=4000× cos (30)
Work done = 4000 × cos 30
Work done= 4000×√3/2
Work done = 2000√3 J
WORK DONE=FORCE × DISPLACEMENT × COS@
WORK=200 ×20× cos 30
Work=4000× cos (30)
Work done = 4000 × cos 30
Work done= 4000×√3/2
Work done = 2000√3 J
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Hey , it's 30° only.... so pls correct your answer :
I will thanks.....
30 or 300?
30
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