C My Portal | Clovis Community E Grades for Kareem Alqatami:F × y WileyPLUS e Ch

Question

C My Portal | Clovis Community E Grades for Kareem Alqatami:F × y WileyPLUS e Chegg Study I Guided Solutio × csecure https://edugen.wileyplus.com/edugen/lti/main.uni Halliday, Fundamentals of Physics, 10e BACK NEXT ASSIGNMENT RESOURCES Chapter 08, Problem 029 Ch. 7, 8) 028 032 013 Your answer is partially correct. Try again. In the figure, a block of mass m = 19 kg is released from rest on a frictionless incline of angle = 33°. Below the block is a spring that can be compressed 1.2 cm by a force of 250 N. The block momentarily stops when it compresses the spring by 5.6 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring? 19 Chapter 0B, Problem 029 Revlew Results by Stud (a) Number (b) Nu Click if you would like to Show Work for this question: UnitsT mm 9.5 Units Open Show Work SHOW HINT LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI LECTURE Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER

Explanation / Answer

Spring constant is k = F/x = 250/0.012 = 20833 N /m

Using law of conservation of energy

Kinetic energy of the block before hitting the spring = Potential energy stored in spring

0.5*m*v^2 = 0.5*k*x^2

0.5*19*v^2 = 0.5*20833*0.056^2

v = 1.85 m/s is the answer for b)

work done by the gravitational Force is W = 0.5*m*v^2

m*g*sin(theta)*S = 0.5*m*v^2

(19*9.8*sin(33)*S) = 0.5*19*1.85^2

S = 0.32 m

so answer for a ) is 0.32 m + 0.056 m = 0.376 m

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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