Question 11, chap 15, seet 4 part 1 of 2 10 points A 1.45 kg block at rest on a

Question

Question 11, chap 15, seet 4 part 1 of 2 10 points A 1.45 kg block at rest on a tabletop is at- tached to a horizontal spring hawing constant 17.7 N/m. The spring is initially unstretched A constant 26.5 N horiaontal force is applied to the object causing the spring to stretch The acceleration of gravity is 9.8m/ Find the speed of the block after it has moved 0.226 m from equilibrium if the coef icient of kinetic friction between block and tabletop is 0.157 Answer in units of m/s 17.7 N/m 5 N 45 k Find the speed of the block after it has oved 0.226 m from equilibeium if the surface between block and tabletop is frietionless Answer in units of m/s Question 12, chap 15, seet 4 part 2 of 2 10 points

Explanation / Answer

11. Applying Work - energy theorem.

Work done by force + Work done by spring = change in KE

(25.5 x 0.226) - (17.7 x 0.226^2 / 2) = 1.45 v^2 /2 - 0

v = 2.71 m/s

12. Now,

Work done by force + Work done by friction + Work done by spring = change in KE

{ f = uk N = 0.157 x 1.45 x 9.8 = 2.23 N }
(25.5 x 0.226) - (17.7 x 0.226^2 / 2) - (2.23 x 0.226) = 1.45 v^2 /2 - 0

v = 2.57 m/s

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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