EXAMPLE 7.12 Work and energy on an air track Video Tutor Solution Now let\'s rev
ID: 2039346 • Letter: E
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EXAMPLE 7.12 Work and energy on an air track Video Tutor Solution Now let's revisit Example 7.8, adding a nonconservative force to the system. Suppose the glider is initially at rest at 0, with the spring unstretched. Then you apply a constant force F with magnitude 0.610 N toO the glider. What is the glider's speed when it has moved tox 0.100 m? SOLUTION Vi SET UP Figure 7.33 shows our sketch. Mechanical energy is not con served because of the work Wother done by the force F, but we can still use the energy relationship of Equation 7.17. Let the initial point be x0 and the final point be x 0.100 m. The energy quantities are Initial F 0.610 N -x,-0.100 m Ki-o, K.-???0.200 kg)of, U=0, 1r=1(5.00 N/m)(0.100 m)2=0.0250 J, Wother = (0.610 N)(0.100 m) = 0.06 10 J V. Final SOLVE Putting the pieces into Equation 7.17, we obtain FIGURE 7.33 Our sketch for this problem. (0.200 kg)u? + 0.0250J000.0610J, mechanical energy isn't conserved, energy considerations simplify this problem greatly 0.600 m/s. REFLECT The total mechanical energy of the system changes by an amount equal to the work Wother done by the applied force. Even though Practice Problem: If the 0.610 N force is removed when the glider reaches the 0.100 m point, at what distance from the starting point does the glider come to rest? Answer: 0.156 mExplanation / Answer
practice problem :
given
m = 0.2 kg, k = 5 N/m
x1 = 0.1 m, v = 0.6 m/s
x2 = ?
let x2 is the final compression of the spring when the blocks comes momentarily rest.
now apply conservation of energy
final mechanical energy = initial mechaincal energy
(1/2)*k*x2^2 = (1/2)*k*x1^2 + (1/2)*m*v^2
x2^2 = x1^2 + (m/k)*v^2
x2 = sqrt(x1^2 + (m/k)*v^2)
= sqrt(0.1^2 + (0.2/5)*0.6^2)
= 0.156 m
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