These questions are Honors physics question. Thank you in an advance

Question

A 120 kg couch is sitting on a marble floor.  When its owner tries to move it, he realizes that it takes him a force of 120 newtons to start it moving.  What is the Static coefficient of friction between the couch and the floor?

2)  A worker on a dock tries to move a crate by pushing with a horizontal force.  If he knows that the coefficient of friction bewtween the crate and the dock is .5, and he needed 500 newtons of force to move it, what does the crate weigh?

3)  If a boy tries to pull his 25 kg brother across a playground with a horizontal force, and the coefficient of friction between the brother and the playground is .75, how much force must the boy apply to move his brother?

4)  A 55 kg crate is sitting on a 35-degree ramp.  What is the minimum static coefficient of friction between the crate and the ramp necessary to keep the crate from sliding down the ramp?

5)  A mouse is tugging on a piece of cheese with a horizontal force of 3 newtons.  If the kinetic coefficient of friction between the cheese and the floor is .25, and the cheese weighs .080 kg, what is the acceleration of the cheese?

6)  A man pulls a 25 kg box of scrap steel across his front yard.  If he was able to effect an acceleration of 5 m/s^2, and he pulled with a horizontal force of 250 newtons, what was the kinetic coefficient of friction between the box and the ground?

7)  Your physics instructor's cat bolts across his carpeted floor to a speed of 10 meters per second.  The cat runs over a section of tiled floor where the kinetic coefficient of friction between its paws and the tile is .15.  Suddenly, the cat decides to stop, to avoid hitting a wall.  If the cat has a mass of 5 kg, and the tiled section is 6 meters long, will the cat stop in time, or will it strike the wall?  In a few words, how far will the cat slide?

8)  A man is pulling his daughter, who is on a sled, across a frozen lake.  The girl and sled have a mass of 45 kg, and the kinetic coefficient of friction between the sled and the ice is .05.  If the rope he is using forms an angle of 40 degrees to the horizontal, and he pulls with a force of 60 newtons, what is the acceleration of the girl and the sled?

9)  Take the same situation as in problem number 7, except that this time, the man is pushing with the same force as before, and the angle is therefore negative 40 degrees.  Find the acceleration of the girl and the sled.

Answer 1

(1)

Force required to move the object F_f is 120 N

Mass of the couch is 120 kg.

F_f  = µ_static (mg)

120 = µ_static  (120 * 9.8)

µ_static  = 120/(120*9.8)

µ_static = 0.102

Therefore coefficient of static friction between couch and floor is 0.102.

Answer 2

(2)

Force required to move the crate F_f is 500 N

Co-efficient of static friction between crate and dock µ_r is 0.5

F_f  = µ_r (mg)

500 = 0.5 (mg)

mg = 500/(0.5)

Weight = 1000

Therefore weight of the crate is 1000 N.

Answer 3

(3)

Mass of the brother is 25 kg.

Co-efficient of static friction between brother and playground µ_r is 0.75.

F_f  = µ_r (mg)

F_f = 0.75 (25* 9.8)

F_f = 183.75

Force required by the boy to move his brother F_f is 183.75 N.

Answer 4

(4)

The Mass of the crate is 55 kg.

Angle θ is 35°

We know that F_f  = µ_static F_N

From the Figure, F_f  = -mgsinθ

F_N  = -mgcosθ

µ_static  = F_f /F_N

µ_static  = mgsinθ/mgcosθ

µ_static  = tanθ

µ_static  = tan(35)

µ_static  = 0.7

Therefore the coefficient of static Friction is 0.7.

Answer 5

(5)

Force required by mouse in tugging a cheese F_app is 3 N

Co-efficient of Kinectic friction between cheese and floor µ_k is 0.25

The Mass of the cheese is 0.80 kg

We know that F_net = ma

ma = F_app - F_f

ma  = F_app - µ_r (mg)

a  = (F_app - µ_r (mg))/m

a = (3 - (0.25*0.80*9.8))/0.80

a = (3-1.96)/0.80

a = 1.04/(0.80)

Accerlation = 1.3

Therefore Accerlation of the cheese is 1.3 m/sec².

Answer 6

(6)

Mass of the scrap steelr is 25 kg.

Accerlation of the scrap steel is 5 m/sec²

Horizantal Force required to move the scrap steel F_f is 250 N

We know that F_net = ma

ma = F_app - F_f

F_f = F_app - ma

µ_k (mg) = F_app - ma

µ_k  = (F_app - ma)/(mg)

µ_k  = (250 - (25*5))/(25*9.8)

µ_k = (250-125)/(25*9.8)

µ_k = 125/245

Coefficient of kinectic friction is µ_k = 0.510

Therefore oefficient of kinectic friction µ_k is 0.51.

Answer 7

(8)

The Coefficient of Kinetic Friction between sled and ice is 0.05.

Mass of the girl and the sled is 45 kg.

Force required to pull the girl with sled is 60 N

Angle at which the rope is used to pull is 40°.

F_app = mg + F sinθ

F_app = 45*9.8+ 60 * sin40

F_app = 441 + 38.56

F_app = 479.56

 

We know that F_net = ma

ma = F_app - F_f

ma  = F_app - µ_r (mg)

a  = (F_app - µ_r (mg))/m

a = (476 - (0.05*45*9.8))/45

a = (476-22.05)/45

a = 453.95/(45)

Accerlation = 10.0877

Therefore Accerlation of the girl with sled is 10 .0877 m/sec².