43.8k views
0 votes
A 22.8 kg rocking chair begins to slide across the carpet when the push reaches 57.0 N. What is the coefficient of static friction?​

A 22.8 kg rocking chair begins to slide across the carpet when the push reaches 57.0 N-example-1

2 Answers

3 votes

Answer:

.255

Step-by-step explanation:

I'm also on acellus and it's the right answer

User John Krueger
by
6.0k points
4 votes

Answer:

0.255

Step-by-step explanation:

The following data were obtained from the question:

Force (F) = 57 N

Mass (m) = 22.8 Kg

Coefficient of static friction (µ) =...?

Next, we shall determine the normal reaction (R). This is illustrated below:

Mass (m) = 22.8 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal reaction (R) =?

R = mg

R = 22.8 x 9.8

R = 223.44 N

Finally, we can obtain the coefficient of static friction (µ) as follow:

Force (F) = 57 N

Normal reaction (R) = 223.44 N

Coefficient of static friction (µ) =...?

F = µR

57 = µ x 223.44

Divide both side by 223.44

µ = 57/223.44

µ = 0.255

Therefore, the coefficient of static friction (µ) is 0.255.

User WhatWhat
by
5.6k points