Figure gives the acceleration of a 2.0kg body as it moves from rest along x-axis while a variable force acts on it from x=0 to x=9 m. The work done by the body when it reaches
1) x=4m
2) x=7m shall be as given below
Answers
answer : option (4) 42 J and 30 J respectively
area under acceleration - position graph multiply with mass of body gives workdone by the body.
so, workdone by the body when it reaches( x = 4m), W = mass of body × area under acceleration - position graph
= 2kg × [area of trapezium ]
= 2kg × 1/2 (4m + 3m) × 6m/s²
= 42 J
again, workdone by the body when it reaches (x = 7m) W' = 2kg × [area of trapezium 0 to 5m - area of trapezium 5 to 7 m]
= 2kg [ 1/2 × (5m + 3m) × 6m/s² - 1/2 × (2m + 1m) × 6m/s²]
= 2kg [ 24m²/s² - 9 m/s² ]
= 30 J
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Answer:(4) 42 joule and 30 joule respectively.
Explanation:
