Answer to Question #125269 in Mechanics | Relativity for Chuka ogbonnaya

Question

• A 50.0 kg circus acrobat drops from a height of 2.0 m
straight down onto a springboard with a force constant of
8.00 × 103 N/m. By what maximum distance does she
compress the spring?
• Using energy considerations and assuming negligible air
resistance, show that a rock thrown from a bridge 20.0 m
above water with an initial speed of 15.0 m/s strikes the
water with a speed of 24.8 m/s independent of the direction
thrown.

Answer 1

By energy conservation, maximum compression occurs when the gravitational potential energy completely stored in the spring system, thus

$$mgh=\frac{1}{2}kx_{max}^2\\ \implies x_{max}=\sqrt{\dfrac{2mgh}{k}}$$

Hence,

$$x_{max}=\sqrt{\dfrac{2\cdot 50\cdot 10\cdot 2}{8000}}=0.5m$$

Note: We assumed $g=10m/s^2$ for simplicity in calculation.

First of all we already knew that energy is a scalar quantity, thus while we calculate the kinetic energy ,it does not matter in which direction the object is thrown.

Now, as air resistance is negligible ,thus on applying the energy conservation we get,

$$mgh+\frac{1}{2}mu^2=\frac{1}{2}mv^2\\ \implies v= \sqrt{2gh+u^2}=\sqrt{2\cdot 9.8\cdot 20 +225}=24.8m/s$$

Note: In this case $g=9.8m/s^2$ .