- #1
moenste
- 711
- 12
Homework Statement
A compressed spring is used to propel a ball-bearing along a track which contains a circular loop of radius 0.10 m in a vertical plane. The spring obeys Hooke's law and requires a force of 0.20 N to compress it to 1.00 mm.
(a) The spring is compressed by 30 mm. Calculate the energy stored in the spring.
(b) A ball-bearing of mass 0.025 kg is placed against the end of the spring which is then released. Calculate (i) the speed with which the ball-bearing leaves the spring, (ii) the speed of the ball at the top of the loop, (iii) the force exerted on the ball by the track at the top of the loop.
Assume that the effects of friction can be ignored.
Answers: (a) 9 * 10-2 J, (b) (i) 2.7 m s-1, (ii) 1.8 m s-1, (iii) 0.55 N
2. The attempt at a solution
(a) E = 0.5 k x2 → E = 0.5 * [(0.2 * 30) / 0.03] * 0.032 = 0.09 J
(b) (i) KE = 0.5 m v2 → v = √(2 KE) / m = √(2 * 0.09) / 0.025 = 2.7 m s-1
(b) (ii) v = √2 g h = √2 * 10 * 0.2 = 2 m s-1 (in this case I took h as 0.1 + 0.1 -- twice the radius)
or v = √g r = √10 * 0.1 = 1 m s-1
or v = √2 g h = √2 * 10 * 0.1 = 1.4 m s-1
or PE = m g h = 0.025 * 10 * 0.1 = 0.025 J → v = √(2 KE) / m = √(2 * 0.025) / 0.025 = 1.4 m s-1
or PE = m g h = 0.025 * 10 * 0.2 = 0.05 J → v = √(2 KE) / m = √(2 * 0.05) / 0.025 = 2 m s-1
So part (b) (ii) is wrong and I can't see my mistake.
(b) (iii) I took v = 1.8 m s-1 from the answer. ∑ F = N + m g = (m v2) / r → N = [(m v2) / r] - m g = [(0.025 * 1.82) / 0.1] - 0.025 * 10 = 0.56 N
As far as I can see only (b) (ii) is wrong. Any suggestions please on where is the mistake? Thanks in advance.