Answer the physics question 3 and 4

Step 1: Convert given values to SI units for Question 3. Mass $m = 20 \text{g} = 20 \times 10^{-3} \text{kg} = 0.02 \text{kg}$ Extension $x = 4 \text{cm} = 4 \times 10^{-2} \text{m} = 0.04 \text{m}$ Force constant $k = 200 \text{Nm}^{-1}$ Step 2: Apply the principle of conservation of energy for Question 3. The elastic potential energy stored in the rubber is converted into the kinetic energy of the stone. $$E_p = E_k$$ $$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$$ Step 3: Substitute the values and solve for velocity $v$ for Question 3. $$\frac{1}{2}(200 \text{ Nm}^{-1})(0.04 \text{ m})^2 = \frac{1}{2}(0.02 \text{ kg})v^2$$ $$(200)(0.0016) = (0.02)v^2$$ $$0.32 = 0.02v^2$$ $$v^2 = \frac{0.32}{0.02}$$ $$v^2 = 16$$ $$v = \sqrt{16}$$ $$v = 4 \text{ m/s}$$ The velocity with which the stone leaves the catapult is $\boxed{\text{4 m/s}}$. Step 4: Identify given values for Question 4. Compression $x = 0.02 \text{m}$ Force constant $k = 400 \text{Nm}^{-1}$ Step 5: Calculate the energy stored in the spring for Question 4. The energy stored in a spring is given by the elastic potential energy formula: $$E_p = \frac{1}{2}kx^2$$ $$E_p = \frac{1}{2}(400 \text{ Nm}^{-1})(0.02 \text{ m})^2$$ $$E_p = \frac{1}{2}(400)(0.0004)$$ $$E_p = 200(0.0004)$$ $$E_p = 0.08 \text{ J}$$ The energy stored in the spring is $\boxed{\text{0.08 J}}$. Send me the next one 📸