<p>A wooden block with mass 2.00 kg is placed against a compressed spring at the

Question

<p>A wooden block with mass 2.00 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 33.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 7.45 m up the incline from A, the block is moving up the incline at a speed of 5.30 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is &#956;k=0.400. The mass of the spring is negligible.<br /><br />* Calculate the amount of potential energy that was initially stored in the spring.<br />Take free fall acceleration to be g= 9.80 m/s^2 .</p>

Explanation / Answer

Solved a similar question already.. with different numbers.. i put in the question and the answer.. hope it helps :) A wooden block with mass 1.10 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 34.0 degrees (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 7.55 m up the incline from A, the block is moving up the incline at a speed of 5.90 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is mu_k = 0.50. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring. Take free fall acceleration to be 9.80 m/s^2. ANSWER : Usual 1st step in solving physics probs: SKETCH the situation showing incline & angle with block's distance, mass and weight and show all forces that act perpendicular and parallel to incline 2nd step: solve for: 1) block's GPE at B and 2) block's KE at B and 3) energy lost to friction {force} = ff x 7.55 {ff = friction force} 3rd step: set SPE = value of: 1) ans + 2) ans + 3) ans --------------------- GPE of block at B = mgh = 1.10(9.80)(7.55 sin 34) = 45.512 J KE of block at B = 1/2mV² = (0.5)(1.10)(5.90)² = 19.146 Normal force of incline against block = 1.10(9.80)cos 34 = 8.9370 ff = friction force = (0.50)(8.9370) = 4.4685 energy loss to friction = (4.4685)(7.55) = 33.737 J SPE = 45.512 + 19.146 + 33.737 = 98.4 J ANS

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.