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Question 3 A person pulls on a 9.3 kg crate as it slides across a level floor using a rope attached to the center of the crate. The floor has a non-zero coefficient of kinetic friction. The crate began with a speed of 1.1 m/s and speeded up to 2.8 m/s while being pulled a horizontal distance of 2.0 meters. If the force exerted by the rope is 20 Newtons, and the rope pulls completely horizontally, what is the work in J done by the friction force between the crate and floor? Question 4 A person pulls on a 7.5 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.9 m/s and speeded up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate?

          Question 3

A person pulls on a 9.3 kg crate as it slides across a level floor using a rope attached to the center of the crate. The floor has a non-zero coefficient of kinetic friction. The crate began with a speed of 1.1 m/s and speeded up to 2.8 m/s while being pulled a horizontal distance of 2.0 meters. If the force exerted by the rope is 20 Newtons, and the rope pulls completely horizontally, what is the work in J done by the friction force between the crate and floor?

Question 4

A person pulls on a 7.5 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.9 m/s and speeded up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate?

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Question 3

A person pulls on a 9.3 kg crate as it slides across a level floor using a rope attached to the center of the crate. The floor has a non-zero coefficient of kinetic friction. The crate began with a speed of 1.1 m/s and speeded up to 2.8 m/s while being pulled a horizontal distance of 2.0 meters. If the force exerted by the rope is 20 Newtons, and the rope pulls completely horizontally, what is the work in J done by the friction force between the crate and floor?

Question 4

A person pulls on a 7.5 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.9 m/s and speeded up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate?

Added by Kathryn D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ravindra Yadav

03/30/2022

Step 1

8 \, m/s\), \(u = 1.1 \, m/s\), and \(d = 2.0 \, m\). Show more…

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A person pulls on a 9.3 kg crate as it slides across a level floor using a rope attached to the center of the crate. The floor has a non-zero coefficient of kinetic friction. The crate began with a speed of 1.1 m/s and sped up to 2.8 m/s while being pulled a horizontal distance of 2.0 meters. If the force exerted by the rope is 20 Newtons and the rope pulls completely horizontally, what is the work in J done by the friction force between the crate and floor? Question 4: A person pulls on a 7.5 kg crate against a 22 Newton frictional force using a rope attached to the center of the crate. The crate began with a speed of 1.9 m/s and sped up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work done by the force applied by the rope on the crate?

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Transcript

00:01 Hello everyone, we are going to understand this question here given in the question.

00:10 Mass of carrot, mass of carrot that is represented by m, which is 9 .3 kg, 9 .3 kilogram.

00:31 Initial speed of carrot, initial speed of carrot, that is represented by u, which is 1 .1 meter per second.

00:53 Final speed, final speed that is represented by v, which is 2 .8 meter per second.

01:09 Distance, distance pulled by it, distance pulled, that is represented by d, which is 2 .0 meter.

01:29 Force exerted by rope, force exerted by rope, that is reaffirited by rope, that is re -percented.

01:44 Presented by f, which is 20 newton.

01:51 Work done by frictional force we have to find.

01:55 Work done by friction force.

02:04 Work done by friction force, which we have to find.

02:14 Let the acceleration be a.

02:18 Let the acceleration be a.

02:31 Now applying, let the acceleration be a, now applying third equation of motion, that is v square.

02:42 Is equal to u square plus 2 into a into d so we can write a is equal to v square minus u square upon 2d now substituting the value in this so we can write a is equal to value of v is 2 .8 square minus value of u that is 1 .1 square upon 2 into value of d is 2 meter per second square.

03:21 Now doing further calculation we will get 6 .63 upon 4 meter per second square...

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