Happy New Year to all.  The last batch of math problems is physics themed and taken from Physics for Scientists and Engineers.  The image is reproduced here.

Problem description:

Given the mass of m1 is 7kg and the mass of m2 is 5kg, calculate the forward acceleration of m1.  The coefficient of kinetic friction between m1 and the surface is 0.54.  Assume the mass m1 is already moving.  (Disregard static friction.)

Solution:

We begin by calculating the force exerted on m2.  Force = mass * acceleration.  We know acceleration due to gravity is 9.81 m/s2.  F_m2 = 5kg * 9.81m/s^2 leaves us with 49.05N.  Now we should calculate the force on m1.  For starters, we need to calculate the force that m1 exerts on the table.  This will give us the surface normal force which, when combined with the coefficient of friction, will provide the ‘resistance’ to forward motion.  F_m1 = 7kg * 9.81m/s^2 = 68.67N.  Multiply by the coefficient of friction and we end up with 0.54*68.67N = 37.0818N.  Again, this 37N force will act against the pull from m2.  49.05N – 37.0818N = 11.9682N.  Now that we have the sum of forces on m1, we can calculate the acceleration.  F = ma.  11.9682N = 7kg * a.  11.9682N/7kg = 1.709742857 m/s^2

Other problems were thematically similar, and dealt with determining the coefficient of sliding friction and calculating the velocity required to keep water in a bucket as it spun in circles.

Progress:

• 3/1000 math problems.
• 0/100 pictures.
• 0/6 games.