Asked by ATT
As a 2.0-kg object moves from (2i+5j)m to (6i-2j)m, the constant resultant force acting on it is equal to (4i-3j)N. If the speed of the object at the initial position is 4.0 m/s, what is its kinetic energy at its final position? Ans: 40J
Answers
Answered by
Anonymous
x distance = 4
y distance = -7
x force = 4
y force = -3
work done on it = 4*4 + -7*-3= 16 + 21 = 37 Joules = increase in Ke
initial Ke = (1/2)(2)(16) = 16 Joules
total energy = 37 + 16 = 53 Joules
y distance = -7
x force = 4
y force = -3
work done on it = 4*4 + -7*-3= 16 + 21 = 37 Joules = increase in Ke
initial Ke = (1/2)(2)(16) = 16 Joules
total energy = 37 + 16 = 53 Joules
Answered by
ATT
your answer is not correct
Answered by
anonymous
53 J is the correct answer
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