1. An object has a mass of 8.0 kilograms. A 2.-newton force displaces the object a distance of 3.0 meters to the east, and then 4.0 meters to the north. What is the total work done on the object?(adsbygoogle = window.adsbygoogle || []).push({});

1)10 J 2) 14 J 3) 28 J 4) 56 J

I found the resultant of the two distances, which is 5 m, I used 5 m * 2 N=10 J, but it’s not the answer.

2. A 20-newton block is at rest at the bottom of a frictionless incline, since I can’t show you the diagram, I will describe it. Supposed a right triangle, the vertical height is 3 m, and the base is 4.0 m. Then the hypotenuse is the incline and the block is at the bottom. The question is: How much work must be done against gravity to move the block to the top of the incline?

I know that conservative forces are forces for which the work done does not depend on the path taken but only on the initial and final positions. But here I Know I should use 3 m * 20N=60 J. but the thing is it starts from the bottom which is the bottom of the incline, and ends at the top, then it should be the length of the hypotenuse? But if it’s 3 m, then it should have started from left end of the base, not the right end of the base which is the bottom of the incline.

Thanks a lot for help.

**Physics Forums - The Fusion of Science and Community**

# Conservative forces

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Conservative forces

Loading...

**Physics Forums - The Fusion of Science and Community**