# hw10b - HW#10b Note numbers used in solution steps are...

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HW#10b Note: numbers used in solution steps are different from your WebAssign values. Page 1 of 6 a). For Spring 2, Spring force: F=k* x, x=0.6-0.1 =0.5 (m), Spring Force = object weight = 3 (N) k = F/ x =3/0.5= 6 (N/m) b.) For Spring 3, F= k* x, x=1.1-0.1 =1.0 (m), Object weight= Spring Force= k* x= 6* 1=6 (N) In orbit, he has centripetal acceleration, so the total net force along the radius direction pointing to the earth center is NOT zero. Hence if he stands on a scale, the normal force from the scale is not equal to mg. Also when he is in orbit, g is NOT equal to 9.8 m/s 2 a). A mass on spring: Oscillation period: , So: So: =3000*1 2 / (4*3.14 2 )=76 kg b). Potential energy stored in the spring U =1/2* k* x 2 When it is Furthest from the equilibrium length: x=Amplitude=4m U =1/2* k* Α 2 =0.5*3000*4 2 =24000 J c). When it passes equilibrium: All Potential energy is converted into Kinetic energy: ½*m*V 2 =U max = 1/2* k* Α 2 =24000 J V 2 =k*A 2 /m d). oscillation period is in dependant to amplitude. It is only determined by the stiffness of the spring (spring constant) and the mass of the object. So T is still 1 s.

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HW#10b Note: numbers used in solution steps are different from your WebAssign values. Page 2 of 6 Problem 3. The height of a tower is measured by attaching a simple pendulum to its ceiling, whose length is barely enough to stay off the floor. The pendulum is let go from a small angle, and takes 12 s to return to the same location it started from. a.) How tall is the tower? m If the pendulum mass is let go 0.2 m above the floor, b.) how fast is the mass traveling as it grazes the floor? m/s Solution. (a) The period of the pendulum is related to the length of the pendulum, H=L
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hw10b - HW#10b Note numbers used in solution steps are...

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