A 1.6 kg breadbox on a frictionless incline of angle theta = 42 degree is connec

Question

A 1.6 kg breadbox on a frictionless incline of angle theta = 42 degree 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? Number _____________ Units Number ____________ Units Number ________________ Units

Explanation / Answer

a)

L = length of incline travelled = x = stretch in the spring = 10.3 cm = 0.103 m

h = height dropped vertically = L Sin42 = 0.103 Sin42

m = mass = 1.6 kg

v = speed gained

using conservation of energy

Potential energy of block = kinetic energy + spring potential energy

mgh = (0.5) m v2 + (0.5) k x2

(1.6) (9.8) (0.103 Sin42) = (0.5) (1.6) v2 + (0.5) (120) (0.103 Sin42)2

v = 0.997 m/s

b)

x' = stretch in the spring when the block stops

T = tension force in the rope

at rest, weight component of block parallel to incline is balanced by the tension force in the rope

hence T = mg Sin42

also along horizontal direction for the spring

T = spring force = k x'

hence k x' = mg Sin42

(120) x' = (1.6) (9.8) Sin42

x' = 0.087 m = 8.7 cm

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