Learning Goal: Part B-The acceleration of the collar after it has moved a certai

Question

Learning Goal: Part B-The acceleration of the collar after it has moved a certain distance To set up and solve the equations of motion using rectangular coordinates What is the acceleration of the collar after it has moved 2.5 m? The 1.5-kg collar shown has a coefficient of kinetic friction Express your answer to three significant figures 0.215 with the shaft. The spring is unstretched when View Available Hint(s) 8-0 and the collar is given an initial velocity of to = 11 m/s. The unstretched length of the spring is d 1.5 m and the spring constant is k 35.3 N/m. (Figure 1) a,--30.3 m/s Figure 1 of 1 > Submit Previous Answers Incorrect, Try Again; 5 attempts remaining Part C . The speed of the collar after it has moved a certain distance What is the speed of the collar after it has moved 2.5 m? Express your answer to three significant figures and include the appropriate units View Available Hint(s)

Explanation / Answer

from the given data

m = 1.5 kg

u = 0.215

vo = 11 m/s

d = 1.5 m

k = 35.3 N/m

b. after the collar has moved x = 2.5 m

spring extension dx = sqroot(d^2 + x^2) - d = 1.415475 m

now, let normal reacotn be N

then friciton f = uN

hence form force balance

a = f/m + kdx*x/sqroot(x^2 + d^2)/m

and

N = mg + k*dx*d/sqroot(x^2 + d^2)

hence

a = u(mg + k*dx*d/sqroot(x^2 + d^2))/m + kx*dx/sqroot(x^2 + d^2)/m

a = -34.3468722 m/s/s

c. work done by friction = dW

dW = u(mg + k*dx'*d/sqroot(x^2 + d^2))dx

integrating

W = u(mgx + k*dx'*d*ln(sqrot(1 + x^2/d^2) + x/d))

W = -24.683086 J

hence speed at this point = v

from energy balance

0.5*m*vo^2 = 0.5*m*v^2 - w + 0.5*k*dx^2

v = 5.305828 m/s

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