A 2.0-kg particle has an initial velocity of (Si-4) m/s, Sometime later, its velocity is 7i...

Question

Question

Answer 1

Answer #1

work done = change in KE

W = $\Delta$ KE

W = 1/2 mv_f² - 1/2 mv_i²

v_f = √(7²+3²) = √58

v_i = √(5²+4²) = √41

W = (1/2 )( 2kg) (58-41) = 17 J (a)

Answer 2

Answer #2

we have m= 2kg and v_i= 5i - 4j and v_f= 7i + 3j

W = △KE = 1/2 m△v²

W = 1/2 mvf2 - 1/2 mvi2

vf = √((7)2+(3)2) = √58

vi = √((5)2+(-4)2) = √41

W = (1/2 )( 2 )(√58)² - (1/2)(2)(√41)²= 17 J

answer = 17 J (a)

answered by: Sarah Hazem

Answer 3

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