A 200 g block is pushed against a horizontal spring of constant 200 N/m until it

Question

A 200 g block is pushed against a horizontal spring of constant 200 N/m until it is compressed 15 cm. When the mass is loses contact with the spring it moves over a horizontal rough patch of surface of length 50 cm and coefficient of friction 0.2. After passing this patch the block continues to slide up a frictionless ramp of angle 30 degree. a) How far does the block rise on the ramp after the first pass? b) The mass comes down and goes over the rough patch again on the way back to the spring. How many passes will the mass make over the patch before running out of energy? c) Where along the rough patch will the block finally stop? d) How long would the patch have to be so that the block never reaches the ramp?

Explanation / Answer

let the block rises to a moves a distance d after first pass

height h = d*sin30

work done = change in KE

Wg + We + Wf = dKE

(U1g-U2g) + (U1e - U2e) + Wf = 0

(0-m*g*h) + ( (1/2)*k*x^2 - 0 ) - uk*m*g*L = 0

-0.2*9.8*d*sin30 + ((1/2)*200*0.15^2) - (0.2*0.2*9.8*0.5) = 0

h = 2.1 m <<<===========ANSWER

===============

(b)

during each pass the mass loses energy = Wf = uk*m*g*L

number of passes n = (1/2)*k*x^2/(uk*m*g*L)

n = ((1/2)*200*0.15^2)/(0.2*0.2*9.8*0.5) = 11.5

number of passes = 12

======================

(c)

after 11 th pass energy left = (1/2)*k*x^2 - 11*Wf

E11 = (1/2)*200*0.15^2 - (11*0.2*0.2*9.8*0.5) = 0.094 J

during the 12th pass the block comes to rest

the block moves from left to right during 12 th pass

let the block stops at point x1 from left

0.094 = 0.2*0.2*9.8*(0.5-x1)

x1 = 0.26 m <<<<---------answer

======================

if total initial energy = Wf

(1/2)*k*x^2 = uk*m*g*L1

(1/2)*200*0.15^2 = 0.2*0.2*9.8*L1

L1 = 5.74 m <<<<-------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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