A 2.3 kg breadbox on a frictionless incline of angle = 44 is connected, by a cor

Question

A 2.3 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 = 100 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 10.4 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

The figure is not given, however I am trying to solve the problem.

(a) Using energy
0.5*100*108.16/100^2+0.5*2.3*v^2=
2.3*9.8*10.4/(100*sin(44))

=> 0.5408 + 1.15*v^2 = 3.374

=> v = 1.57 m/s

(b) When the energy in the spring is equal to the loss of PE, the box stops

0.5*k*x^2=m*g*x/sin(44)
solve for x
x=2*m*g/(k*sin(44)) = 0.65 m

(c) and (d) The acceleration is upward equal to the force of the spring minus the force of gravity parallel to the incline divided by the mass

a=(100*0.65-2.3*9.81*sin(44))/2.3 = 21.45 m/s^2

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