A 2.5 kg breadbox on a frictionless incline of angle ? = 37 ? is connected, by a

Question

A 2.5 kg breadbox on a frictionless incline of angle ? = 37 ? 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 10.1 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 displacement along the incline = 0.101m

The potential energy lost by the box =mgh sin
2.5*9.8* 0.101*sin 37 = 1.49 J
The kinetic energy gained by the box = 0.5mv^2 = 0.5*2.5*v^2 = 1.25 v^2 J
Work done in stretching the spring or
The potential energy gained by the spring =0.5 kx^2
= 0.5*100*0.101^2 = 0.51 J
Equating the energies
1.49 =1.25 v^2 + 0.51
Solving for v
v = 0.885m/s
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b)
As before
2.5*9.8* d *sin 37 = 0 + 0.5*100*d^2

d = 0.295m
---------------------------------------...
c

The upward force exerted by the spring at this position is
kx = 100*0.295m = 29.5N
The downward pulling force is mg sin 37 =14.74N
The net force acting up = 29.5-14.74 = 14.76 N


The upward acceleration is force /mass = 14.76 /2.5 = 5.89 m/s^2
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