LuanaVrbas367
Answered

Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of kinetic friction between the box and the floor?0.11 a)0.10 b)0.13 c)0.09 d)0.11

Sagot :

AL2006

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero. 

Since you're pushing on it with 53N in that direction, friction must be pulling
on it with 53N in the other direction.

 The 53N of friction is (the weight) x (the coefficient of kinetic friction).

                                                  53N  =  (490N) x (coefficient).

Divide each side by  490N :  Coefficient = (53N) / (490N)  =  0.1082 .

Rounded to the nearest hundredth, that's    0.11 .      (choice 'd')
The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.
If the box is sliding at constant speed, and not speeding up or slowing down, that means that the horizontal forces on it add up to zero.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.