LuanaVrbas367
Answered

At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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.
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.