A 25.0 kg child on a 1.00 m long swing is released from rest when the ropesof th

Question

A 25.0 kg child on a 1.00 m long swing is released from rest when the ropesof the swing make an angle of 25.0° with the vertical. (a) Neglecting friction, find the child's speedat the lowest position.
1 m/s

(b) If the speed of the child at the lowest position is1.20 m/s, what is the mechanical energylost due to friction?
2 J (a) Neglecting friction, find the child's speedat the lowest position.
1 m/s

(b) If the speed of the child at the lowest position is1.20 m/s, what is the mechanical energylost due to friction?
2 J

Explanation / Answer

Part a: A = (1.00 m)(sin25) = .423 y = Asin25 = .423sin25 = .179 m = height of swing PE = KE mgh = 1/2 mv^2 gh = 1/2 v^2 (9.8)(.179) = 1/2 v^2 v = 1.87 m/s Part b: KE = 1/2mv^2 = 1/2 (25 kg)(1.87 m/s)^2 = 43.8 J (If none islost) KEreal = 1/2mv^2 = 1/2(25 kg)(1.20 m/s)^2 = 18 J Lost due to friction: 43.8 - 18 = 25.8 J

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