A 1.7 kg breadbox on a frictionless incline of angle ? = 42 ? is connected, by a
ID: 1429728 • Letter: A
Question
A 1.7 kg breadbox on a frictionless incline of angle ? = 42 ? 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 10.3 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
P.E of box=Ub=-mgh
=-mg(dsin42)
P.E of spring=Us=1/2kd^2
K.E of box
Ki=0
Kf=1/2mvf^2
kf+Us+Uf=0
1/2mvf^2+1/2k
d^2-mgdsin 42=0
rearranging we get
vf=undrt(2/m(mgdsin 42-1/2kd^2)
=undrt(2dsin42-1/2kd^2/m)
=undrt(2*0.103*sin 42-0.5*120*0.103^2/1.7)
=undrt(0.137-0.374)
vf=0.486m/sec--answer a
b)mgdsin 42=1/2kd^2
d=2mgsin 42/k
=2*1.7*9.8sin 42/120
d=0.185m--answer b
c)The upward force e
kx = 120*0.103m = 12.36N
The downward pulling force is mg sin 42 =11.14N
The net force acting up = 12.36-11.14 = 1.22 N
The upward acceleration is force /mass = 1.22/1.7 = 0.717 m/s^2
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