Question

A Stone of mass 40kg is released from

Answer 1

Answer:

v = 3.53x10⁻³ m/s.

Explanation:

We can find the velocity of the rock by energy conservation:

$\Sigma E_{i} = \Sigma E_{f}$        

$\frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}$

Where:

i is for initial and f is for final

k: is the stiffness of the rubber = 20 N/m

x: is the distance of the rubber compression = 5 mm = 0.005 m

m: is the mass of the stone = 40 kg

v: is the velocity =?

Hence, the velocity is:

$v = \sqrt{\frac{kx^{2}}{m}} = \sqrt{\frac{20 N/m(0.005 m)^{2}}{40 kg}} = 3.53 \cdot 10^{-3} m/s$

Therefore, the velocity with which the stone leaves the catapult is 3.53x10⁻³ m/s.

 

I hope it helps you!