0 webassign.net 0 ?72%ND Tue 1.13 pMJae Gubri eE Q Webassign.ne.. I chegg com 0
ID: 2038943 • Letter: 0
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0 webassign.net 0 ?72%ND Tue 1.13 pMJae Gubri eE Q Webassign.ne.. I chegg com 0 magnitude equal to: Phy 1010-Ch 13 Hom-Wns 1FI (3 N/m)(3.00 x 102 m)1-k- (3 N/m)(3.00 x 102 m)- 0.09 (B) At what value of x is the force from the spring equal to 44% of the maximum force? Set l F 1-1-kx l i equal to 44% of the maximum value, without keeping track of the minus sign because the question sis about the mageude ortheforce From this it is seen that the force is reduced to 44% of its maximum magnitude where Irl has 44% of its maximum value, or at x cm (C At what value of x is the speed of the block equal to 44% of its maximum speed? we can answer this question from the solutions for the posit of time. Because the block is at its maximum displacement x-re at time r-O, the equation for rs: and veeory ass ons (1) x cos(or) with ?-0. Therefore the equation for the velocity is: (2) v =-ano sin(or) whose maximum magnitude is ano. The time t when the speed is 44% of this manimum value therefore satisfies: (3) 100 aar?-ane, Isin(ari This is satisfied for: sinter]-100 But because sin'(aar) +cos(ear)-1, Equation (4) reduces to: And therefore the locations were the speed has the required values are: c em ±0.898Explanation / Answer
part B:
maximum force=0.09 N
given, at x such that
|k*x|=0.44*0.09
==>k*x=0.0396 or -0.0396
using k=3 N/m,
x=0.0132 m or x=-0.0132 m
hence answer is :
x=1.32 cm or x=-1.32 cm
part C:
equation for x=x0*cos(w*t)
where x0=amplitude=3 cm
then speed=v=dx/dt=-x0*w*sin(w*t)
maximum value of speed=x0*w as maximum value of (-sin(w*t)) is 1.
so let at time t, speed is 44% of maximum speed
==>0.44*x0*w=|-x0*w*sin(w*t)|
==>sin(w*t)=0.44 or sin(w*t)=-0.44
hence cos(w*t)=sqrt(1-0.44^2) or -sqrt(1-0.44^2)
=0.898 or -0.898
then value of x=x0*cos(w*t)
=2.694 cm or -2.694 cm
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