A 1.6 kg breadbox on a frictionless incline of angle theta = 41 degree is connec

Question

A 1.6 kg breadbox on a frictionless incline of angle theta = 41 degree is connected, by a cord that runs over a pulley, to a light spring of spring constant k = 110 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. What is the speed of the box when it has moved 10.2 cm down the incline? How far down the incline from its point of release does the box slide before momentarily stopping, and what are the magnitude and direction of the box's acceleration at the instant the box momentarily stops?

Explanation / Answer

(a) Let "d" be the distance that the box moves along the slope. The vertical height distance "h" that the box moves is:

h = d*Sin()

The change in gravitational potential energy equals the accumulation of spring potential energy plus the kinetic energy acquired by the box:

m*g*h = (1/2)*k*(d^2) + (1/2)*m*v^2

Multiply by two, cancel the "m", substitute in the "h" derived above, and rearrange:

2*g*d*Sin() - (k/m)*d^2 = v^2

v = sqrt( 2*9.8*.102*Sin(41) - (110/1.6)*.102^2 ) = 0.77m/s .

(b)m*g*h = (1/2)*k*(d^2) + (1/2)*m*v^2

v=0

2*g*d*Sin() = (k/m)*d^2

One "d" cancels:

d = (m/k)*2*g*Sin() = ( 1.6/110)*2*9.8*Sin( 41 ) = 0.187 m

(c The force of the spring is: F = k*d , parallel to the slope. The component of force due to gravity, parallel to the slope is: m*g*Sin(). The net force is:

F = k*d - m*g*Sin()

a = F/m = k*d/m - g*Sin() = 110*.187/1.6 - 9.8*Sin(41) = 6.43 m/sec^2

d. The acceleration is parallel to the incline surface, and up.

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