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

Step-by-step 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* 20* 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* 20* 9.8=43.12\ N`

net Force on crate
`F-F_k=ma`

`a=(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* 7.84* 2`

`v=√(31.37)`

`v=5.60\ m/s`