Alternative Exercise 7.123 What is the speed of the package just before it reach

Question

Alternative Exercise 7.123 What is the speed of the package just before it reaches the spring? A 2.00-kg package is released on a 53.1° incline 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline (Figure 1) The coefficients of friction between the package and the incline are , = 0.40 and k = 0.20. The mass of the spring is negligible m/s Submit My Answers Give Up Part B What is the maximum compression of the spring? Figure 1 of 1 m = 2.00 kg Submit My Answers Give Up 4.00 m Part C The package rebounds back up the incline. How close does it get to its initial position? 53·1° Submit My Answers Give Up

Explanation / Answer

Given,

m = 2 kg ; theta = 53.1 deg ; d = 4 m ; us = 0.4 and uk = 0.2

a)Let v be the speed.

From conservation of energy

KE = PE - Wf

where Wf is the work by by the friction

PE = m g h = m g d sin(theta)

Wf = uk m g d cos(theta) ; KE = 1/2 m v^2

1/2 m v^2 = m g d sin(theta) - uk m g d cos(theta)

1/2 v^2 = dg [sin(theta - uk cos(theta)]

v = sqrt {2 g d [sin(theta - uk cos(theta)]}

v = sqrt [2 x 9.81 x 4 (sin53.1 - 0.2 x cos53.1)] = 7.3 m/s

Hence, v = 7.3 m/s

B)We know that the potential enery of the spring is:

U = 1/2 k x^2

from conservation of energy:

KE + U - KE - PE = W

1/2 k x^2 - x [m g sin(theta) - uk m g cos(theta)] + [uk m g d cos(theta) - m g d sin(theta)] = 0

0.5 x 120 x x^2 - x (2 x 9.8 x sin53.1 - 0.2 x 2 x 9.8 x cos53.1) + (0.2 x 2 x 9.8 x 4 cos53.1 - 2 x 9.8 x 4 sin(53.1) ] = 0

60 x^2 - 13.32 x - 53 = 0

the abive quadratic eqn gives us:

x = 1.06 m , -0.837

Hence, x = 1.06 m

c)again from conservation of energy

1/2 k x^2 + m g x sin(theta) + m g D sin(theta) = -uk m g x cos (theta) + uk m g D cos(theta)

D = k x^2/2[mg sin(theta) + uk mg cos(theta)] - x

D = 120 x 1.06^2/ 2 x 9.8 x 2 (sin53.1 + 0.2 x cos53.1) - 1.06 = 2.68 m

Do the distance down the incline is:

X = 4 - 2.68 = 1.32 m

Hence, X = 1.32 m

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