A horizontal spring having constant k is attached to a block having mass M. The mass is oscillating freely on a frictionless table with amplitude A. As the mass travels from the position of the maximum stretch to half the position of maximum stretch, how much work is done by the spring on the mass?
Given Data
k = 50.0 N/m
M= 0.50 kg
A= 15.0 cm
xₒ=A
x=A/2
W=?
Solution
U=½kx²
Uₒ=½kxₒ²
Uₒ-U=W
W=½kxₒ²-½kx²=½k(xₒ²-x²)
W=½k(A²-(½A)²)=½k(A²-(½A)²)=½k(A²-¼A²)=½k(¾A²)=⅜kA²
W=⅜·(50 N/m)·(0.15m)²
Code for the Google Calculator: 3/8*(50 N/m)*(0.15m)^2
0.421875 joules
Answer: W=0.42J
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