A 1.5 kg breadbox on a frictionless incline of angle theta = 36 degree is connec

Question

A 1.5 kg breadbox on a frictionless incline of angle theta = 36 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. (a) What Is the speed of the box when it has moved 10.6 cm down the incline? (b) How far down the incline from its point of release docs 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? (a) Number Unit (b) Number Unit (c) Number unit (d)

Explanation / Answer

here ,

theta = 36 degree

m = 1.5 Kg

k = 110 N/m

a) x = 10.6 cm = 0.106 m

let the speed of breadbox is v

Using conservation of energy

0.5 * m * v^2 = m * g * x * sin(theta) - 0.5 * k * x^2

0.5 * 1.5 * v^2 = 1.5 * 9.8 * 0.106 * sin(36) - 0.5 * 110 * 0.106^2

solving for v

v = 0.63 m/s

the speed of box is 0.63 m/s

b) for the box to stop

Using conservation of energy

0.5 * m * v^2 = m * g * x * sin(theta) - 0.5 * k * x^2

0 = 1.5 * 9.8 * x * sin(36) - 0.5 * 110 * x^2

solving ffor x

x = 0.158 m

the block stops after travelling 0.158 m

c) acceleration of block = (k * x - m * g * sin(theta))/m

acceleration of block = (110 * 0.158 - 1.5 * 9.8 * sin(36))/1.5

solving

acceleration of block = 5.83 m/s^2

the acceleration of the block is 5.83 m/s^2

d)

the direction of motion of the acceleration is upwards

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