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A 1.3-kg mass is attached to a spring with a force constant of52 N>m. If the mass is released with a speed of 0.28 m>s at a distance of 8.1 cm from the equilibrium position of the spring, whatis its speed when it is halfway to the equilibrium position? |
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Answer» GIVEN info : A 1.3-kg mass is attached to a spring with a force constant of 52 N/m. If the mass is released with a speed of 0.28 m/s at a distance of 8.1 cm from the EQUILIBRIUM position of the spring. To find : the speed when it is HALFWAY to the equilibrium position, is... solution : initial velocity, u = 0.28 m/s initial position, x₁ = 8.1 cm = 0.081 m final position, x₂ = x₁/2 = 0.0405 m using energy conservation theorem, initial KINETIC energy + initial potential energy = final kinetic energy + final potential energy ⇒1/2 mu² + 1/2 kx₁² = 1/2 mv² + 1/2 kx₂² ⇒m(v² - u²) = k(x₁² - x₂²) ⇒1.3(v² - 0.28²) = 52(0.081² - 0.0405²) ⇒1.3(v² - 0.0784) = 0.256 ⇒v² - 0.0784 ≈ 0.2 ⇒v² = 0.2784 ⇒v ≈ 0.53 m/s THEREFORE the speed of mass is 0.53 m/s |
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