Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the chair is in motion, a 327 N horizontal force keeps it moving at a constant velocity.

a)Find the coefficient of static friction between the chair and the floor.
b)Find the coefficient of kinetic friction between the chair and the floor.

Sagot :

Answer:

Here's what you know, m=25 kg, Ff=365 N, g=9.8 m/s^2, use the formula Fg = mg, Fg =25*9.8, Fg =245 N, then use the formula Ff <= ตsFn, 365=245ตs, solve for ตs,  ตs=1.5.

Here's what ya know, m=25 kg, Ff=327 N, the normal force of gravity will be the same as in part A.  Now use the formula Ff = µkFn, 327=245 µk, solve for µk, µk=1.3.

Explanation:

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.