A wooden block with mass 1.90 kg is placed against a compressed spring at the bo
ID: 1543743 • Letter: A
Question
A wooden block with mass 1.90 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 34.0 (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 5.55 m up the incline from A, the block is moving up the incline at a speed of 5.90 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is k = 0.55. The mass of the spring is negligible.
A.
Calculate the amount of potential energy that was initially stored in the spring.
Take free fall acceleration to be 9.80 m/s2 .
Explanation / Answer
initial elastic potential energy = Ui
final eleastic potential eenrgy Uf = 0
work done by frictional force Wf = -k*m*g*costheta*d
work done by elastic force We = Ui
work done by gravitational force Wg = m*g*sintheta*d
initial KE Ki = 0
final kE = Kf - (1/2)*m*v^2
from work energy relation
total work = change in KE
ui + m*g*sintheta*d - uk*m*g*costheta*d = (1/2)*m*v^2
Ui + (1.9*9.8*sin34*5.55) - (0.55*1.9*9.8*cos34*5.55) = (1/2)*1.9*5.9^2
Ui = 22.4 J <<<<----------answer
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