Question

X Incorrect The initially stationary 25-kg block is subjected to the time-varying force whose magnitude P is shown in the plot. Note that the force is zero for all times greater than 5 s. Determine the time ts at which the block comes to rest. P, N P 212 25 kg - 34 - Ik = 0.47 Mg = 0.64 0 0 5 t, s 3.119 Answe: ts s the tolerance is +/-2%

Answer 1

Step 1

Equation of applied force, P = ct

212 at t = 5s, P = 212 N = 212 = c(5) c= = 5

212 ..P=3 5 ......... (1)

Step - 2

Now first find time(tm) at which block starts moving

PSinayo AN PCos 34° MEN u E 28(9:81)

balancing vertical direction, N = 25(9.81)-P sin 34° = 245.25-P sin 34°

$Block\,\,will\,\,start\,\,moving\,\,just\,\,after\,\,P\cos34\degree\,\,became\,\,equal\,\,to\,\,static\,\,friction(\mu _sN)$

$P\cos34\degree=\mu _sN=0.64(245.25-P\sin34\degree)$

$\Rightarrow P\approx132.24\,N$

132.24 212 Ftm tm = 5 132.24(5) 212 3.12 s

Step 3

$Now\,\,after\,\,time\,\,\bold{t_m}\,\,block\,\,starts\,\,moving,so\,\,now\,\,kinetic\,\,friction\,\,will\,\,now\,\,act\,\,as\,\,shown$

$FBD\,\,of\,\,block\,\,from\,\,t=t_m=3.12\,s\,\,to\,\,t=5\,s$

A PSinyo AN SPCosyo YN 23(981)

Balancing vertical force, N = 25(9.81) - P sin 34° 245.25 – P sin 34°

$Balancing\,\,horizontal\,\,force,P\cos34\degree-\mu _kN=25(a)$

$P\cos34\degree-0.47(245.25-P\sin34\degree)=25(a)$

$Putting\,\,value\,\,P\,\,from\,\,(1)\,\,and\,\,a=\frac{dV}{dt}$

$\frac{212}{5}t(\cos34\degree+0.47\sin34\degree)-115.2675=25\frac{\mathrm{d} V}{\mathrm{d} t}$

$46.294t-115.2675=25\frac{\mathrm{d} V}{\mathrm{d} t}$

$25\int_{0}^{V}dV=\int_{3.12}^{5}(46.294t-115.2675)dt$

$25V=\left [23.147t^2-115.2675 \right ]_{3.12}^{5}$

$\boldsymbol{V=5.466\,m/s}$

$So\,\,after\,\,5\,\,second\,\,Velocity\,\,of\,\,block,V=5.466\,m/s$

$\boldsymbol{Step-4}$

$FBD\,\,of\,\,block\,\,after\,\,t=5\,s$

z 5 ao MEN 2015.81

$Balancing\,\,vertical\,\,force,N=25(9.81)=245.25\,N$

$Balancing\,\,horizontal\,\,force,-\mu _kN=25(a_o)\Rightarrow a_o=-4.6107\,m/s^2$

$\mathrm{so\,\,after\,\,t=5\,s,block\,\,moving\,\,with\,\,V=5.466\,m/s\,\,will\,\,accelerate\,\,with\,\,a_o=-46107\,m/s^2\,\,and\,\,stop\,\,after\,\,t=t_s}$$\therefore using\,\,equation,0=V+a_o(t_s-5)\Rightarrow \boldsymbol{t_s=\frac{5.466}{4.6107}+5=6.19\,s}$