# As shown in figure 1, a thin string is suspended form a ceiling. When a weight (mass: 500 g)is attached to the lower end, it descends 10 cm and come to rest, as shown in figure 2. Next, as shown in...

As shown in figure 1, a thin string is suspended form a ceiling. When a weight (mass: 500 g)is attached to the lower end, it descends 10 cm and come to rest, as shown in figure 2.

Next, as shown in figure 3, the weight is pulled down 20cm, where it touches the floor. The string's restoring force is proportional to the string's extension from its natural length. The restoring force is not exerted when the string is at less than its natural length.

When the weight is gently released from its position in figure 3, what maximum height (from the floor) does it attain?

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### 1 Answer

In the second figure, since the weight spontaneosly came to rest, it is in equilibrium. This means the restoring force is balanced out by the force of gravity.

Since the restoring force is proportional to the string's extension, it is equal `kl` , where l is the extension (10 cm = 0. 1 m) and k is the constant of proportionality, which is the property of the string.

The gravity force is `mg` , where m is the mass of the weight (500g = 0.5 kg) and the g is the free fall acceleration, which we can approximate to be 10 m/s^2.

In equilibrium, the magnitues of the forces are equal:

`kl = mg`

From here we can find constant k:

`k = (mg)/l =(0.5 kg * 10 (m/s^2))/0.1m = 50 N/m`

The maximum height the weight will reach when it is gently released (not given any initial speed) from the position in figure 3 is given by the energy conservation: all potential energy of the streched string will convert into gravitational energy of the weight.

Potential energy of the streched string is `U_(strech) = (kL^2)/2`

Here L is the extenstion of the string in figure 3, L = 20 cm+ 10 cm = 30 cm = 0.3 m

Potential energy of the weight at the height h from the floor is

`U_(weight) = mgh`

Let's assume that the string is no longer streched when the weight reaches the maximum height. Then the potential energy of the string will be 0 when the weight is at maximum height. If the height h will turn out to be greater than 30 cm, our assumption will prove to be correct.

Under this assumption, `(kL^2)/2 = mgh`

Solving for h, `h=(kL^2)/(2mg)=(50*(0.3)^2)/(2*0.5*10) = 0.45m = 45 cm`

The maximum height is 45 cm. The length of the string is less than its natural length, so there is no strech potential energy. The assumption was correct.

**The maximum height from the floor attained by the weight is 45 cm.**

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