tj MasteringPhysice HW 011 . Energy Part 2-Google Chrome Secure https://session.

Question

tj MasteringPhysice HW 011 . Energy Part 2-Google Chrome Secure https://session.masteringphysics ew?assignmentProblemiD-816255718 offset next Hw #11-Energy Part 2 Problem 6.56 Constants 1 Periodic Table Part A A 280-g wood block is firmly aftached to a very light horizontal spring, as shown in the figure (Figure 1). The block can slide along a table where the coefficient of friction is 0.30 A force of 22 N compresses the spring 18 cm maximum extension? If the spring is released from this position, how far beyond its equilbrium position williR stretich at its frst Express your answer using two significant figures Figure 1of1 Submit Provide Feedback

Explanation / Answer

Solution :- Given data

Mass of blook = 0.280g

Coefficient of friction = 0.30

Force = 20N

F = k*x
22 = k*0.18
k =122.22 N/m

Now, applying the energy conservation equations, we have,
0.5*122.22*0.182 - 0.3*0.28*9.8*x = 0.5*122.22*x2

1.98 - 0.82*x = 61.11*x2

Solving the quadratic, we get, x = 0.173 m

Hence, the distance from which it differes from its equilibrium position = 0.18- 0.173 = 0.007m = 0.7 cm Ans

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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