A 2.00-kg package is released on a 53.1° incline, L = 4.02 m from a long spring

Question

A 2.00-kg package is released on a 53.1° incline, L = 4.02 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline (see figure below). The coefficients of friction between the package and the incline are µs = 0.40 and µk = 0.15. The mass of the spring is negligible.

(a) What is the speed of the package just before it reaches the spring?

m/s

b) What is the maximum compression of the spring?

m

(c) The package rebounds back up the incline. How close does it get to its initial position?

m

Explanation / Answer

intial potential energy = m*g*L*sin53.1

frictional force f = uk*m*g*cos53.1

work done by the frictional force before compressing the spring = Wf = f*(L) = uk*m*g*cos53.1*L

KE of the package = KE = 0.5*m*v^2

from work energy theorem

W = change in PE

uk*m*g*cos53.1*L = m*g*L*sin53.1 - 0.5*m*v^2

v = sqrt(2*((g*Lsin53.1) - (uk*g*L*cos53.1))))

v = sqrt(2*((9.8*4.02*sin53.1) - (0.15*9.8*4.02*cos53.1))))

v = 7.477 m/s   <--------answer

work done by the frictional force after compressing the spring = Wf = f*(L+x) = uk*m*g*cos53.1*L + uk*m*g*cos53.1*x

elastic potential energy stored = 0.5*k*x^2

from work energy theorem

W = change inpotential energy

uk*m*g*cos53.1*L + uk*m*g*cos53.1*x = m*g*L*sin53.1 - 0.5*k*x^2

(0.15*2*9.8*cos53.1*4.02)+(0.15*2*9.8*cos53.1*x) = (2*9.8*4.02*sin53.1) - (0.5*120*x^2)

7.096 + 1.765*x = 63.009 - 60*x^2

x = 0.951 m    <----------------answer

aftger travelling s distance the package stops

0.5*k*x^2 - m*g*(s+x)*sin53.1= uk*m*g*cos53.1*(s+x)

(0.5*120*0.951^2) - (2*9.8*(s+0.951)) = 0.15*2*9.8*cos53.1*(s+0.951)

s = 1.589 m

closeness = L-s = 4.02 - 1.589 = 2.431 m <----------answer

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