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A light elastic string has natural length $2a$ and modulus of elasticity $4mg$ . One end of the elastic string is attached to a fixed point $O$ . A particle $P$ of mass $m$ is attached to the other end of the elastic string. The particle $P$ hangs freely in equilibrium at the point $E$ , which is vertically below $O$ (a) Find the length $OE$ . Particle $P$ is now pulled vertically downwards to the point $4mg$ $0$ , where $4mg$ $1$ , and released from rest. The resistance to the motion of $P$ is a constant force of magnitude $4mg$ $3$ . (b) Find, in terms of $4mg$ $4$ and $4mg$ $5$ , the speed of $P$ after it has moved a distance $4mg$ $4$ . Particle $P$ is now held at $O$ Particle $P$ is released from rest and reaches its maximum speed at the point $O$ $1$ . The resistance to the motion of $P$ is again a constant force of magnitude $4mg$ $3$ . (c) Find the distance $O$ $4$ .

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Q. A light elastic string has natural length

2a

and modulus of elasticity

4mg

. One end of the elastic string is attached to a fixed point

O

. A particle

P

of mass

m

is attached to the other end of the elastic string. The particle

P

hangs freely in equilibrium at the point

E

, which is vertically below

O

(a) Find the length

OE

. Particle

P

is now pulled vertically downwards to the point

4mg

0

, where

4mg

1

, and released from rest. The resistance to the motion of

P

is a constant force of magnitude

4mg

3

. (b) Find, in terms of

4mg

4

and

4mg

5

, the speed of

P

after it has moved a distance

4mg

4

. Particle

P

is now held at

O

Particle

P

is released from rest and reaches its maximum speed at the point

O

1

. The resistance to the motion of

P

is again a constant force of magnitude

4mg

3

. (c) Find the distance

O

4

Equilibrium of P: (a) Since P hangs freely in equilibrium, the tension in the string equals the weight of P. $\newline$ Tension, $T = \text{weight of P} = mg$ $\newline$ Modulus of elasticity, $E = 4mg$ $\newline$ Natural length of string, $L_0 = 2a$ $\newline$ Extension, $e = OE - L_0$ $\newline$ Since $T = E \times (e / L_0)$ , we have $mg = 4mg \times (e / 2a)$ $\newline$ Solve for $e$ : $e = (mg / 4mg) \times 2a = (1/4) \times 2a = a/2$ $\newline$ Length $OE = L_0 + e = 2a + a/2 = 5a/2$
Motion of P at A: (b) When P is pulled to A and released, it moves with a constant resistance $R = (1/4)mg$ . Work done against resistance when P moves a distance $a$ : $W = R \cdot a = (1/4)mg \cdot a$ Initial potential energy at A, $U = mgh$ , where $h = 2a$ (extension from natural length) Kinetic energy at A, $KE = 0$ (since it starts from rest) Final potential energy after moving a distance $a$ , $U' = mg(h - a) = mg(2a - a) = mga$ Using energy conservation, initial energy = final energy + work done against resistance $mgh = (1/2)mv^2 + mga + (1/4)mg \cdot a$ $2mga = (1/2)mv^2 + (5/4)mga$ Solve for $a$ $0$ : $a$ $1$ $a$ $2$ $a$ $3$
Acceleration of P: (c) When P is released from O, it accelerates due to gravity and the resistance. $\newline$ Net force on P, $F = mg - R = mg - \frac{1}{4}mg = \frac{3}{4}mg$ $\newline$ Using $F = ma$ , we get $a = \frac{F}{m} = \frac{3}{4}g$ $\newline$ Maximum speed occurs when kinetic energy is maximum, which is when potential energy change equals work done against resistance. $\newline$ Let $OB = x$ , then the extension is $x - 2a$ . $\newline$ Potential energy change, $\Delta U = mg(x - 2a)$ $\newline$ Work done against resistance, $W = R \times x = \frac{1}{4}mg \times x$ $\newline$ Using energy conservation, $\Delta U = KE_{\text{max}} + W$ $\newline$ $mg(x - 2a) = \frac{1}{2}mv_{\text{max}}^2 + \frac{1}{4}mg \times x$ $\newline$ Solve for $x$ : $F = ma$ $0$ $\newline$ But $F = ma$ $1$ (from part b, with $F = ma$ $2$ replaced by $x$ ) $\newline$ $F = ma$ $4$ $\newline$ $F = ma$ $5$ $\newline$ $F = ma$ $6$ $\newline$ $F = ma$ $7$ $\newline$ $F = ma$ $8$ $\newline$ $x = \frac{$8$}{$3$}a