Answer to Question #290839 in Molecular Physics | Thermodynamics for dandan

Question #290839

The block in figure reaches a velocity of 12m/s in 30m, starting from rest

Compute the coefficient kinetic friction between the block and the ground.


  • 718 N
  • P = 268 N
1
Expert's answer
2022-01-26T17:46:28-0500

Force (p)=268N

Weight (W)=718

Newton second law

Force in vertical direction

(Fnet)y=may(F_{net})_y=ma_y

WN=mayW-N=ma_y

ay=0a_y=0

W=NW=N

Newton motion second law

v2=u2+2asv^2=u^2+2as

Put value

122=02+2×a×3012^2=0^2+2\times a \times30

ax=1442×30=2.4m/sec2a_x=\frac{144}{2\times30}=2.4m/sec^2

Force in horizontal direction

PF=maxP-F=ma_x

718μkN=max718-\mu_kN=ma_x

m=Wg=2689.8m=2689.8=27.34kgm=\frac{W}{g}=\frac{268}{9.8}\\m=\frac{268}{9.8}=27.34kg

Put value

718μk×268=27.34×2.4718-\mu_k\times268=27.34\times2.4

718μk×268=65.616μk×268=652.384μk=2.434718-\mu_k\times268=65.616\\\mu_k\times268=652.384\\\mu_k=2.434


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