Q: A body of mass 100g is attached to a hanging spring whose force constant is 10 N/m. The body is lifted until a the spring is in its unstrached state and then released. Calculate the speed of the body when it strikes the table 15 cm below the release point

(a) 1 m/s

(b) 0.866 m/s

(c) 0.225 m/s

(d) 1.5 m/s

Ans: (b)

Sol:

$ \displaystyle 0.1 \times 10 \times 0.1 = \frac{1}{2}\times 10 \times (0.15)^2 + \frac{1}{2}\times 0.1 \times v^2$