Text Solution Answer : A::B Solution : Here, `u = 30km//h = (30xx(5)/(18))m//s = (25)/(3)m//s (1km//h = (5)/(18) m//s)` <br> `upsilon = 60km//h = (60xx(5)/(18))m//s= (50)/(3)m//s` <br> m = 1500kg <br> According to work-energy theorem, <br> `W = (1)/(2)m upsilon^2 - (1)/(2)mu^2 = (1)/(2)m(upsilon^2 - u^2)` <br> or `W =(1)/(2)xx1500kg[((50)/(3)m//s)^2 - ((25)/(3)m//s)^2]` <br> `=750kg[((2500)/(9)-(625)/(9))(m//s)^2]` <br> `=750kgxx208xx208.33(m//s)^2 = 156250J` Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now |