A 2.4 kg breadbox on a frictionless incline of angle = 44 is connected, by a cor

Question

A 2.4 kg breadbox on a frictionless incline of angle = 44 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. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10.8 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 (d) direction of the box's acceleration at the instant the box momentarily stops?

Explanation / Answer

Given,

m = 2.4 kg ; theta = 44 deg ; k = 120 N/m ;

a)d = 10.8 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 2.4 x 9.81 x 0.108 x sin44 - 120 x 0.108^2]/2.4 = 0.943 m/s

Hence, v = 0.943 m/s

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 2.4 x 9.81 x sin44/120 = 0.273 m

Hence, x = 0.273 m = 27.3 cm

c)considering the forces in y direction

ma = k x - mg sin(theta)

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

a = (120 x 0.273 - 2.4 x 9.81 x sin44]2.4 = 6.84 m/s^2

Hence, a = 6.84 m/s^2

d)Upwards.

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