2) Suppose you have a 120.0-kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood surfaces. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the crate starts to slip, what will its acceleration then be? The coefficient of sliding friction is known to be 0.300 for this situation

Question

contestada

Respuesta :

Khoso123 Khoso123 · Answer 1 · 2020-03-17T04:01:06+01:00

Answer:

[tex](a)F_{max}=588N\\(b)acceleration=1.96m/s^2[/tex]

Explanation:

Given data

Mass m=120.0 kg

Coefficient of static friction [tex]u_{k}=0.500[/tex]

The coefficient of sliding friction [tex]u_{d}=0.300[/tex]

For Part (a) Maximum force

According to Newtons second law the net force aced on the body is given by

[tex]F_{net}=ma=F_{max}-f_{stat}=0\\so\\F_{max}=f_{stat}[/tex]

The friction force is given by

[tex]f_{stat}=u_{stat}*N\\f_{stat}=0.5*(9.8*120)\\f_{stat}=588N[/tex]

Conclude that

[tex]F_{max}=f_{stat}=588N[/tex]

For Part (b) Acceleration

The acceleration due to dynamic friction is given by:

[tex]ma=F_{max}-f_{dyn}[/tex]

The dynamic friction is given by:

[tex]f_{dyn}=u_{dyn}*N\\f_{dyn}=0.300*(9.8*120)\\f_{dyn}=353N[/tex]

So the acceleration given by

[tex]a=\frac{F_{max}-f_{dyn}}{m}\\ a=\frac{588N-353N}{120kg}\\ a=1.96m/s^2[/tex]