Question

A block of mass 14.2 kg is sliding at an initial velocity of 6.15 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.153. (Indicate the direction with the signs of your answers.)

(a) What is the force of kinetic friction (in N) acting on the block?___ N

(b) What is the block's acceleration (in m/s²)? ___m/s²

Answer 1

Solution.

$m=14.2kg;$

$v_0=6.15 m/s;$

$\mu=0.153;$

a)$F_f=\mu N=\mu mg;$

$F_{fx}=-\mu mg;$

$F_{fx}=-0.153\sdot14.2\sdot10=-21.726N$ ;

b) $F{fx}=ma \implies a=\dfrac{F_fx}{m};$

$a=-\dfrac{\mu mg}{m}=-\mu g;$

$a=-0.153\sdot10=-1.53m/s^2;$

Answer: $a) -21.726N;$

$b)-1.53 m/s^2.$