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A 0.60 kg bob hangs on a string 1.5 m long.

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A $0.60\ \text{kg}$ bob hangs on a string $1.5\ \text{m}$ long. It is pulled aside until the string makes $50^\circ$ with the vertical, and let go. Taking $g = 9.8\ \text{m/s}^2$ , find

its speed at the lowest point;
the tension in the string there.

Need a reminder? The key ideaShow

When only conservative forces (gravity, springs) do work, mechanical energy is conserved: $K_1 + U_1 = K_2 + U_2$ . Forces that do no work, like a smooth surface's push or a string at right angles to the motion, do not spoil it.

On a curved path, use energy for the speed and $F = \dfrac{mv^2}{r}$ toward the centre for the forces. At the top of a loop the track's push $N$ and the weight both point down: $N + mg = \dfrac{mv^2}{R}$ , and the cart stays on while $N \ge 0$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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