In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.6 kg book is pushed from rest through a distance of 0.90 m by the horizontal 22 N force from the broom and then has a speed of 1.60 m/s, what is the coefficient of kinetic friction between the book and floor?
Workin= workdoneonfriction+ final KE
22*.9= mu*3.6*g + final KE
solve for mu.
Ok, so I did...
(22)(.9)=mu*(3.6)(9.8)+4.608
19.8=mu(35.28)+4.608
15.192=35.28*mu
mu=0.4306
This is the wrong answer so what am I doing wrong??
u r dumb stop posting you idiot
I apologize if my previous response was not helpful. Let's go through the problem step by step to find the mistake in your calculations.
First, we need to calculate the work done on the book by the horizontal force applied by the broom. The formula for work is:
Work = Force * Distance
So, the work done on the book is:
Work = 22 N * 0.90 m = 19.8 J
Next, we need to calculate the final kinetic energy (KE) of the book. The formula for kinetic energy is:
KE = (1/2) * mass * velocity^2
Plugging in the given values, the final KE of the book is:
KE = (1/2) * 3.6 kg * (1.60 m/s)^2 = 5.76 J
Now, according to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. So, we can write:
Work = Change in KE
19.8 J = (5.76 J - Initial KE)
To find the initial KE, we need to find the initial speed of the book. Since the book starts from rest and is pushed by the broom, the initial speed is 0 m/s.
Initial KE = (1/2) * mass * (initial velocity)^2 = 0 J
Now we can rewrite the equation:
19.8 J = (5.76 J - 0 J)
Simplifying, we get:
19.8 J = 5.76 J
This implies that the work done by the broom is equal to the change in kinetic energy, which means there is no friction acting on the book.
Therefore, the coefficient of kinetic friction should be zero, not 0.4306 as you calculated.
I apologize for any misunderstanding or confusion caused by my previous response.