A 1.00 -kg block is resting against a light, compressed spring at the bottom of a rough plane inclined at an angle of \(30.0^{\circ}\); the coefficient of kinetic friction between block and plane is \(\mu_{\mathrm{k}}=0.100 .\) Suppose the spring is compressed \(10.0 \mathrm{~cm}\) from its equilibrium length. The spring is then released, and the block separates from the spring and slides up the incline a distance of only \(2.00 \mathrm{~cm}\) beyond the spring's normal length before it stops. Determine a) the change in total mechanical energy of the system and b) the spring constant \(k\).

Initially, the block is at the bottom of the incline, so its gravitational potential energy is 0 J.

Initially, the block is at rest, so its kinetic energy is also 0 J.

Since the spring is compressed 10 cm (\(0.1\,\text{m}\)), its elastic potential energy is given by: $$E_s = \frac{1}{2}kx^2$$ The initial energy state of the system is: $$E_i = 0\,\text{J} + 0\,\text{J} + \frac{1}{2}k(0.1^2)$$ #Step 2: Calculate the final energy state of the system# When the block stops, it doesn't have any kinetic energy. We need to find the gravitational potential energy and the elastic potential energy of the spring.

The vertical height (h) at the stopping point is given by: $$h = (0.1 + 0.02) \sin{(30^\circ)} = 0.06\,\text{m}$$ So, the gravitational potential energy is: $$PE = mgh = (1.00\,\text{kg})(9.8\,\text{m/s}^2)(0.06\,\text{m}) = 0.588\,\text{J}$$

At the stopping point, the spring is at its equilibrium length, so there is no elastic potential energy. The final energy state of the system is: $$E_f = 0.588\,\text{J} + 0\,\text{J}$$ #Step 3: Calculate the change in total mechanical energy# Using the work-energy theorem, we can write: $$ΔME = W_{f} = E_f - E_i = 0.588\,\text{J} - \frac{1}{2}k(0.1^2)$$ Now we need to find the work done against friction.

The normal force (N) can be found using: $$N = mg\cos{30^\circ}$$ Now we can find the friction force (f) using the coefficient of kinetic friction: $$f = \mu_{k}N = (0.100)(1.00\,\text{kg})(9.8\,\text{m/s}^2)\cos{30^\circ} = 0.848\,\text{N}$$ The work done against friction is the product of the friction force and the distance the block slides: $$W_{f} = fd = 0.848\,\text{N} \cdot 0.12\,\text{m} = 0.1018\,\text{J}$$ We can now solve for ΔME: $$ΔME = 0.1018\,\text{J}$$ #Step 4: Calculate the spring constant# We can substitute the expression for ΔME into the work-energy theorem equation: $$0.1018\,\text{J} = 0.588\,\text{J} - \frac{1}{2}k(0.1^2)$$ Now we solve for k: $$k = \frac{0.588\,\text{J} - 0.1018\,\text{J}}{0.005\,\text{m}^2} = 97.24\,\text{N/m}$$ So, the spring constant is approximately 97.24 N/m, and the change in total mechanical energy of the system is 0.1018 J.