Question

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the other pulls in the same direction with a force of 380 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Answer 1

0.285

Step-by-step explanation:

Given two forces of different magnitude, it is important to note that the product of normal force and coefficient of kinetic friction should be equal to the sum of these two forces at equilibrium. Therefore, this can be Mathematically expressed as:

`N/\mu_k=F_1+F_2\\\\`

where N is normal force,
`\mu` is coefficient of static friction, F is force and subscripts 1 and 2 represent larger and smaller magnitude forces respectively. Making
`\mu` the subject of the formula then

`\mu_k=(F_1+F_2)/(N)`

Since normal force N is also given by mg where m is mass of object and g is acceleration due to gravity then substituting N with mg we obtain that

`\mu_k=(F_1+F_2)/(mg)` and substituting the figures given in the question, taking g as 9.81 we obtain that

`\mu_k=\frac { 430 N+380 N}{290* 9.81}=0.285`

Hence,the coefficient of kinetic energy is 0.285 as calculated