You push a manual lawn mower across the lawn at constant speed. What is the value of the coefficient of friction between the mower and the grass

Answer:

Explanation:

The question is incomplete. Here is the complete question:

You are pushing a 13.3 kg lawn mower across the lawn with a force of 200 N. What is the value of the coefficient of friction between the mower and the grass if the mower moves with a constant velocity? The force is applied downward at an angle of 65° with the horizontal.

According to Newton's second law of motion:

[tex]\sum F_x= ma_x\\F_{app} - F_f = ma_x\\[/tex]

[tex]F_f = \mu R\\[/tex]

[tex]F_{app} - \mu Rcos \theta = ma_x[/tex]

Fapp is the applied force = 200N

Ff is the frictional force

[tex]\mu[/tex] is the coefficient of friction between the mower and the grass

R is the reaction

m is the mass of the object

ax is the acceleration

Given

R = mg = 13.3*9.8

R = 130.34N

m = 13.3kg

ax  = 0m/s² (constant velocity)

Fapp = 200N

[tex]\theta = 65^0[/tex]

Substitute the given parameters into the formula and get the coefficient of friction as shown;

Recall that: [tex]F_f = \mu R\\[/tex]

[tex]\mu = \frac{F_f}{R}\\\mu = \frac{F_{x}cos65}{F_y+W} \\\mu =\frac{ 200cos65}{200sin65+13.3(9.8)}\\\mu = \frac{84.52}{181.26+130.34}\\\mu = \frac{84.52}{311.6}\\\mu = 0.27[/tex]

Hence the coefficient of friction between the mower and the grass is 0.27

g A 240-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s