When a 0.2 kg mass is suspended from a vertically hanging spring it stretches th

Question

When a 0.2 kg mass is suspended from a vertically hanging spring it stretches the spring from its original length of 0.5 cm to a total length of 6.0 cm. The spring with the same mass attached is then placed on a horizontal frictionless surface. The mass is pulled so that the sprig stretches to a total length of 10 cm. The mass is released and it oscillates back and forth. A/ what is the maximum speed of the mass as it oscillates?

B/ a tennis ball of mass 0.060 kg is served. It strikes the ground with a velocity of 54m/s ( 120 mi/h) at an angle of 220 below the horizontal. Just after the bounce it is moving at 53 m/s at an angle of 180 above the horizontal. If the interaction with ground lasts 0.065s, what average force did the ground exert on the ball?

C/ a 2 kg uniform flat disk is thrown into the air with a linear speed of 10 m/s. As it travel the diak spins at 3.0 rev/s. If the radious of the disk is 10 cm, what is the magnitude of its angular momentum?

Explanation / Answer

solution: m = 0.2 kg dL = 0.060 m - 0.050 m = 0.01 m dL' = 0.1 m k = mg/dL = (0.2)(9.8)/0.01 k = 196 N/m ?F = ma -kx = m d²x/dt² d²x/dt² + (k/m)x = 0 if ?² = k/m d²x/dt² + ?² x = 0 solution for this ordinary differential equation homogen - second order is x(t) = C1 cos ?t + C2 sin ?t d/dt (x(t)) = -C1 ? sin ?t + C2 ? cos ?t C1 and C2 are constantas initial velocity is Vo = d/dt (x(0)) = -C1 ? sin ?(0) + C2 ? cos ?(0) = C2 ? = 0 ? ? 0.................so C2 = 0 initial displacement is x(0) = C1 cos ?(0) + C2 sin ?(0) = C1 = dL' = 0.1 m final solution is x(t) = 0.1 cos ?t general equation if simple harmonic motion is x(t) = A cos ? t these equations are similar, so the conclusion is Amplitude = 0.1 m ? = ? = v(k/m) 2pf = v(k/m) f = 1/(2p) v(k/m) f = 1/(2p) v(196/0.2) f = 4.98 Hz analyze the speed of the block each time the spring is 5.0 cm long x(t) = 0.1 cos ?t 0.05 = 0.1 cos ?t cos ?t = 0.5 for one cycle ?t = p/3 and 5p/3 ? = v(k/m) ? = 31.3 v(t)1 = (d/dt) x(t) = - 0.1 ? sin ?t v(t)1 = - 0.1 (31.3) sin (p/3) v(t)1 = -2.71 m/s v(t)2 = - 0.1 (31.3) sin (5p/3) v(t)2 = 2.71 m/s

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