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

Question

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

Answer 1

Force (p)=268N

Weight (W)=718

Newton second law

Force in vertical direction

$(F_{net})_y=ma_y$

$W-N=ma_y$

$a_y=0$

$W=N$

Newton motion second law

$v^2=u^2+2as$

Put value

$12^2=0^2+2\times a \times30$

$a_x=\frac{144}{2\times30}=2.4m/sec^2$

Force in horizontal direction

$P-F=ma_x$

$718-\mu_kN=ma_x$

$m=\frac{W}{g}=\frac{268}{9.8}\\m=\frac{268}{9.8}=27.34kg$

Put value

$718-\mu_k\times268=27.34\times2.4$

$718-\mu_k\times268=65.616\\\mu_k\times268=652.384\\\mu_k=2.434$