Question

The coefficients of friction between the 20-kg crate and the inclined surface are µ,8 = 0.24 and J.lk = 0.22. If the crate starts from rest and the horizontal force F = 200 N, what is the magnitude of the velocity of the crate when it has moved 2 m?

Answer 1

Answer:5.60 m/s

Explanation:

Given

Coefficient of static friction $\mu _s=0.24$

Coefficient of kinetic friction $\mu _k=0.22$

mass of crate $m=20\ kg$

Force applied $F=200\ N$

maximum static Friction $F_s=\mu _sN$

$N=mg$

$F_s=0.24\times 20\times 9.8$

$F_s=47.04\ N$

thus applied force is greater than Static friction therefore kinetic friction will come into play

$F_k=\mu _kN$

$F_k=0.22\times 20\times 9.8=43.12\ N$

net Force on crate $F-F_k=ma$

$a=\frac{200-43.12}{20}=7.84\ m/s^2$

Magnitude of velocity can be obtained by using

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here initial velocity is zero as crate start from rest

$v^2-0=2\times 7.84\times 2$

$v=\sqrt{31.37}$

$v=5.60\ m/s$