A block of mass 5 kg initially at rest at the origin is acted on by a force along the x-positive direction represented by F = (20 + 5x) N. Calculate the work done by the force during the displacement of the block from x = 0 to x = 4 m.
Answers
Answered by
35
Hii friend,
● Answer- 120 J
● Explaination-
# Given-
m = 5 kg
x1 = 0 cm
x2 = 4 cm
# Solution-
Here, Force = 20 + 5x
Work done is given by,
Work = Force × displacement
W = ∫(20+5x).dx from x=0 to x = 4
W = 20x + 5x^2 /2
W = 20×4 + 5×16/2
W = 120 J
Work done in moving particle from x = 0 to x = 4 cm.
Hope that was useful...
● Answer- 120 J
● Explaination-
# Given-
m = 5 kg
x1 = 0 cm
x2 = 4 cm
# Solution-
Here, Force = 20 + 5x
Work done is given by,
Work = Force × displacement
W = ∫(20+5x).dx from x=0 to x = 4
W = 20x + 5x^2 /2
W = 20×4 + 5×16/2
W = 120 J
Work done in moving particle from x = 0 to x = 4 cm.
Hope that was useful...
Answered by
2
Answer:
120 J
Explanation:
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