Question #238610

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

1
Expert's answer
2021-09-23T08:33:12-0400

Solution.

m=14.2kg;m=14.2kg;

v0=6.15m/s;v_0=6.15 m/s;

μ=0.153;\mu=0.153;

a)Ff=μN=μmg;F_f=\mu N=\mu mg;

Ffx=μmg;F_{fx}=-\mu mg;

Ffx=0.15314.210=21.726NF_{fx}=-0.153\sdot14.2\sdot10=-21.726N ;

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

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

a=0.15310=1.53m/s2;a=-0.153\sdot10=-1.53m/s^2;

Answer: a)21.726N;a) -21.726N;

b)1.53m/s2.b)-1.53 m/s^2.




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