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REALLY NEED HELP ON THESE QUESTIONS WILL MARK BRAINLYEST 100POINTS UP FOR GRABS!!!!!


4. A 4.5 kg box is pushed with a force of 38 N. It accelerates across the floor at a rate of 5.9 m/s2 . Calculate force friction.

5. A student slid a 1.9 kg textbook across a desk. The book’s acceleration was 6.5 m/s2 . The force friction on the table was 5.6 N. What was the magnitude of the force that the student exerted on the book?

6. When a 0.8 kg hockey puck is struck with a 570 N force, it accelerates at a rate of 682.5 m/s2 through the air. What is the wind drag (friction due to the wind)?

7. Which is stronger—static or sliding friction? Justify your answer using examples.

8. A 3.8 kg box is pushed with a force of 27 N. It accelerates across the floor at a rate of 4.8 m/s2 . Calculate force friction.

Respuesta :

#4

when an object is pushed by applied force it accelerated by

a = 5.9 m/s^2

now here we will have

[tex]F_{ap} = 38 N[/tex]

m = 4.5 kg

[tex]a = 5.9 m/s^2[/tex]

now we can say that net force on the object is applied force and friction force

this net force is given by Newton's II law as

[tex]F_{net} = ma[/tex]

[tex]F_{ap} - F_f = ma[/tex]

[tex]38 - F_f = 4.5*5.9[/tex]

[tex]38 - F_f = 26.55[/tex]

[tex]F_f = 11.45 N[/tex]

so friction force will be 11.45 N

#5

when an object is pushed by applied force it accelerated by

a = 6.5 m/s^2

now here we will have

[tex]F_{f} = 5.6 N[/tex]

m = 1.9 kg

[tex]a = 6.5 m/s^2[/tex]

now we can say that net force on the object is applied force and friction force

this net force is given by Newton's II law as

[tex]F_{net} = ma[/tex]

[tex]F_{ap} - F_f = ma[/tex]

[tex]F_{ap} - 5.6 = 1.9*6.5[/tex]

[tex]F_{ap} - 5.6= 12.35[/tex]

[tex]F_{ap} = 17.95 N[/tex]

so applied force will be 17.95 N

#6

Given that

mass of hockey Puck = 0.8 kg

applied force on the puck = 570 N

acceleration of the puck = 682.5 m/s^2

now we can say that net force on the object is applied force and drag force of the air

this net force is given by Newton's II law as

[tex]F_{net} = ma[/tex]

[tex]F_{ap} - F_d = ma[/tex]

[tex]570 - F_d = 0.8*682.5[/tex]

[tex]570 - F_d = 546[/tex]

[tex]F_d = 570 - 546 = 24 N[/tex]

so drag force due to air on the puck will be 24 N

#7)

Static friction is always stronger than sliding friction because in static situation of the object the bonding of the molecules of two surfaces will be very strong and hence it requires more force to start the motion, once the motion has been started the bonding of two surfaces becomes weak and then object can continue its motion with less friction.

#8

Given that

mass of box = 3.8 kg

applied force on the box = 27 N

acceleration of the box = 4.8 m/s^2

now we can say that net force on the object is applied force and friction force of the ground

this net force is given by Newton's II law as

[tex]F_{net} = ma[/tex]

[tex]F_{ap} - F_f = ma[/tex]

[tex]27 - F_f = 3.8*4.8[/tex]

[tex]27 - F_f = 18.24[/tex]

[tex]F_f = 27 - 18.24 = 8.76 N[/tex]

so friction force due to ground on the box will be 8.76 N

Answer:

8.76 is the answer

Explanation:

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