A force of 26.0 N is required to start a 3.2 kg box moving across a horizontal concrete floor. If the 26.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?
I got answers like 2.8, .051, 1.6, and .625
None of these were correct. Please help and explain your answers.
Also, I have posted other such questions. Even if you don't know how to do this one, please check those out in case you know how to do those. Thanks!
2007-12-18 07:54:30 · 2 answers · asked by Bollywood Masti 4 in Science & Mathematics ➔ Physics
FBD of the box
Sum of the horizontal forces = m*a = F - friction
Sum of the vertical forces = 0 = N - m*g ==> N = m*g
friction = u*N = u*m*g
m*a = F - friction = F - u*m*g
solve for u
u = (F-m*a)/(m*g)
Plug and chug
***
I got u = 0.77
2007-12-18 07:59:30 · answer #1 · answered by civil_av8r 7 · 1⤊ 0⤋
The friction force is typically modeled as μmg where μ is the coefficient of friction, m the mass, and g the local acceleration of gravity.
To get it started: 26.0 N = μ ( 3.2 kg ) ( 9.8 m/s² )
Solve for μ, the coefficient of static friction.
Once the box is moving, kinetic friction takes over. The coefficient of kinetic friction is always less than or equal to that of static friction. Now Newton's Second Law is
F = ma
( 3.2 kg ) ( .50 m/s² = 26.0 N - μ ( 3.2 kg ) ( 9.8 m/s² )
Solve for μ, the coefficient of kinetic friction.
2007-12-18 16:02:03 · answer #2 · answered by jgoulden 7 · 0⤊ 0⤋