This problem involves a block sliding down a frictionless incline, then encountering a flat surface with friction, and finally compressing a spring. We'll solve each part step by step:

Given:

• Mass of the block, $m = 10 \, \text{kg}$
• Initial height, $h = 3 \, \text{m}$
• Distance from point B to C, $d = 6 \, \text{m}$
• Coefficient of friction on BC, $\mu = 0.4$
• Spring constant, $k = 2250 \, \text{N/m}$
• Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$

a) Kecepatan benda di titik B

To calculate the velocity at point B, we use the principle of conservation of mechanical energy. Since the track AB is frictionless, all potential energy at point A is converted to kinetic energy at point B.

• Potential energy at point A: $PE_A = mgh = 10 \times 9.8 \times 3 = 294 \, \text{J}$

• Kinetic energy at point B: $KE_B = \frac{1}{2} mv_B^2$

Setting $PE_A = KE_B$, we get: $294 = \frac{1}{2} \times 10 \times v_B^2$ $v_B^2 = \frac{294 \times 2}{10} = 58.8$ $v_B = \sqrt{58.8} \approx 7.67 \, \text{m/s}$

Thus, the velocity at point B is approximately $v_B = 7.67 \, \text{m/s}$.

b) Kecepatan benda di titik C

Between B and C, the block encounters friction, so the kinetic energy at B is reduced by the work done by friction.

• Work done by friction: $W_{\text{friction}} = \mu m g d = 0.4 \times 10 \times 9.8 \times 6 = 235.2 \, \text{J}$

• Kinetic energy at point C: $KE_C = KE_B - W_{\text{friction}} = 294 - 235.2 = 58.8 \, \text{J}$

• Velocity at point C: $KE_C = \frac{1}{2} mv_C^2$ $58.8 = \frac{1}{2} \times 10 \times v_C^2$ $v_C^2 = \frac{58.8 \times 2}{10} = 11.76$ $v_C = \sqrt{11.76} \approx 3.43 \, \text{m/s}$

Thus, the velocity at point C is approximately $v_C = 3.43 \, \text{m/s}$.

c) Pemendekan maksimum dari pegas

At point C, the block compresses the spring. The kinetic energy at C is fully converted into elastic potential energy of the spring.

• Elastic potential energy of the spring: $PE_{\text{spring}} = \frac{1}{2} k x^2$

Setting $KE_C = PE_{\text{spring}}$: $58.8 = \frac{1}{2} \times 2250 \times x^2$ $x^2 = \frac{58.8 \times 2}{2250} = 0.05227$ $x = \sqrt{0.05227} \approx 0.2285 \, \text{m}$

Thus, the maximum compression of the spring is approximately $x = 0.2285 \, \text{m}$.

Final Answers:

a. $v_B = 7.67 \, \text{m/s}$

b. $v_C = 3.43 \, \text{m/s}$

c. $x = 0.2285 \, \text{m}$

Would you like more details or have any questions?

Here are 5 related questions to extend the concept:

1. How would the results change if the incline were not frictionless?
2. What if the mass of the object were doubled?
3. How would the compression of the spring change with a different spring constant?
4. How much energy is lost due to friction between points B and C?
5. How would the system behave if the coefficient of friction increased?

Tip: In problems involving energy conservation, always account for energy losses due to friction when surfaces are not frictionless.