T and Pushing Box

# T and Pushing Box - o F N1S = m*a o 300N*10m/s 2 o 3000 N...

This preview shows pages 1–2. Sign up to view the full content.

25). A fish (8 kg) is being yanked vertically upward out of the water and the fishing line breaks. If the line is rated to a maximum tension of 240 N (~ 53 lb test), then what was the minimum acceleration of the fish (in m/s 2 )? T + W1E = m*a 240kg + -80 N W1E = 8kg * a a= 20 m/s 2 26). A crate (40 kg) is hung in an elevator by a cord rated to withstand a tension of 700 N. The elevator starts at rest and begins moving upwards until the cord holding the crate breaks. What was the acceleration of the elevator when the cord broke? Assume g = 10 m/s 2 . Multiply the weight of the crate to find force in Newtons. o 40kg*10 m/s 2 = 400 N Subtract the force you just calculated from the withstanded tension. This gives you the net force. o 700N – 400N = 300 N Then divide net force by the weight of the crate. o 300 N / 40 kg o 7.5 m/s 2 27). A box (300 kg) is on the floor. The coefficient of static friction = 0.8 between the box and the surface of the shelf. Assume g = 10 m/s 2 . What is the normal force on the box due to the surface of the shelf?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: o F N1S = m*a o 300N*10m/s 2 o 3000 N What is the maximum amount oF Force that can be exerted on the box due to static Friction? o F NIS * μ= F max o 3000 N* 0.8 o 2400 N The box is pushed with a force of 2350.0 N. What is the amount of force exerted on the box due to static friction? o The static friction force is equal to the opposing applied force o F μ =|-2350 N| o 2350 N The static friction will continue to increase until the box is pushed with a force of 2400.0 N. At this point the box will begin to move and therefore the motion of the box will now be opposed by kinetic friction. If the coeFcient of kinetic friction = 0.3, what is the magnitude of the acceleration ? o μ – μ f 0.8-0.3 = 0.5 o Use 0.5 and multply by F N1S 3000 N * 0.5 1500 N o Divide The max amounT of force exerTed based o± of new μ To geT The magniTude of The acceleraton. 1500 N / 300kg 5 m/s 2...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern