I need help solving this problem and understanding. Please show all work so that

Question

I need help solving this problem and understanding. Please show all work so that I can try and understand how this problem works.

Problem #1 (3+1+1+1 6 points) A 5.5 kg block is attached to a massless spring with the spring constant of 220 N/m and is at rest on a horizontal surface with the coefficient of kinetic friction of uk 0.1. The block is struck by a 1 traveling horizontally at 5.5 m/s as shown in the figure. The stone rebounds back at 2 m/s. Note that the friction must be taken into account. Find (a) the speed of the block (in m/s) right after the collision with the stone (1 point) (b) the maximum distance (in m) that the block compresses the spring after the collision (1 point); (c) the work (in J) done by the friction force during this process point.

Explanation / Answer

let

m = 1.1 kg, v1 = 5.5 m/s, v2 = -2m/s
M = 5.5 kg, V1 = 0, V2 = ?

a) apply conservation of momentum

momentum right after the collision = momentum right before the collision

M*V2 + m*v2 = M*V1 + m*v1

5.5*V2 + 1.1*(-2) = 0 + 1.1*5.5

V2 = (1.1*5.5 + 1.1*2)/5.5

= 1.5 m/s <<<<<------------------Answer

b) let x is the maximum distance travelled

workdone by friction = change in mechaincal enery of the block

Fk*x*cos(180) = (1/2)*M*V2^2 - (1/2)*k*x^2

-mue_k*M*g*x = (1/2)*M*V2^2 - (1/2)*k*x^2

-0.1*5.5*9.8*x = (1/2)*5.5*1.5^2 - (1/2)*220*x^2

on solving the above equation we get

x = 0.263 m <<<<<------------------Answer

c) Workdone by friction = Fk*x*cos(180)

= -mue_k*M*g*x

= -0.1*5.5*9.8*0.263

= -1.42 J <<<<<------------------Answer

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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