Question

a A 2.0 kg breadbox on a incline of angle 8 = 30° is connected, by a cord that runs over a pulley, to a light spring of spring constant k = 120 N/m, as shown in the Figure 4. The coefficient of kinetic friction between box and the inclined plane is Mik = 0.52, The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless and the system is isolated. (a) What is the speed of the box when it has moved 10 cm down the incline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (a) direction (up or down the incline) of the boxs acceleration at the instant the box momentarily stops? 000000 e Figure 4: Problem 4.

Answer 1

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m = 2.0 kg ; theta = 30 deg ; k = 120 N/m ;

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(a)

d = 10 cm

We know that,

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

from conservation of energy:

m g d sin(theta) = 1/2 k d^2 + 1/2 m v^2

solving for v we get:

v = sqrt [2 m g d sin(theta) - k d^2]/m

v = sqrt [2 x 2.0 x 9.81 x 0.10 x sin30 - 120 x 0.10^2]/2.0 = 0.617m/s

Hence, v = 0.617 m/s

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(b)

from conservation of energy:

m g x sin(theta) = 1/2 k d^2

solving for d we get:

x = 2 m g sin(theta)/k

x = 2 x 2.0 x 9.81 x sin30/120 = 0.1635 m

Hence, x = 0.1635 m = 16.35 cm

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(c)

considering the forces in y direction

ma = k x - mg sin(theta)

a = [(k x) - mg sin(theta)]/m

a = (120 x 0.1635 - 2.0 x 9.81 x sin30]/2.0 = 4.905 m/s^2

Hence, a = 4.905 m/s^2

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(d)

Upwards.

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