The acceleration may be calculated by using two consecutive intervals. Let t1 be the first time interval of the two consecutive intervals and t2 be the second time interval of the two consecutive time intervals. Let the constant distance interval be represented by s. If the acceleration, a, is constant, then the equations of a uniformly accelerated motion apply. For the first interval the following equation applies:
s = v1i t1 + at1 2 ⁄ 2 . . . . . .(1)
where v1i is the initial velocity of the first interval. And for the second interval the following equation applies:
s = v2i t2 + at2 2 ⁄ 2 . . . . . .(2)
where v2i is the initial velocity of the second interval. The final velocity, v1f, of the first interval is the same as the initial velocity, v2i, of the second interval. Therefore,
v2i = v1f = v1i + at1 . . . . . .(3)
Substituting equation (3) into equation (2) we obtain
s = v1it1t2 + a ( t1t2 + t1 2 ⁄ 2 ) . . . . . ( 4 )
The time intervals, t1 , t2 and the constant distance interval, s, will be measured. Thus, we will have only two unknowns, v1i and a, in equations (1) and (4). That means we can solve for the acceleration a simultaneously in terms of t1 , t2 and s to obtain
a = 2s ( t1 - t2 ) ⁄ [ t1 t2 ( t1 + t2 ) ] . . . . . ( 5 )
That is, if two consecutive time intervals and the constant distance interval are measured, the acceleration can be calculated.
To calculate the coefficient of friction, we will use an arrangement where an object of mass m1 sliding on a table is attached to a hanging object of mass m2 by means of a string passing through a pulley. The forces acting on the object sliding on the table ( m1 ) are the tension in the string which is equal to the weight of the hanging object (Weight of hanging object = m2 |g| ) and the force of friction ( f ) between the two surfaces. The normal pressing force is equal to the weight of the sliding object. (Weight of sliding object = m1 |g| ). Therefore f = μm1 |g|. Since the direction of friction is opposite to the direction of motion, the net force acting on the sliding object is the difference between the tension in the string = and the force of friction. Applying Newton’s second law to the acceleration of these two objects we obtain,
m2 |g| - μm1 |g| = (m1 + m2 ) a
where g is the gravitational acceleration whose value is –9.8m/s2. Now solving for the coefficient of friction ( ), we get,
μ = { m2 |g| - ( m1 + m2 ) a } ⁄ ( m1 |g| )
Therefore if the masses m1 , m2 and the acceleration, a, are measured, the value of the coefficient of friction ( μ ) can be calculated.
And if an experiment is designed so that the motion is uniform (i.e, a = 0 ), the coefficient of friction may be calculated by setting the acceleration to zero in the above equation:
μ = m2 ⁄ m1
An approximate uniform motion can be obtained by finding a mass of the hanging object for which after the sliding object is given a little push, the sliding object moves with a uniform speed in the immediate local area. (Only the immediate local area should be considered because the value of the coefficient of friction may vary from place to place slightly.)
hanger mass = kg = 0.005 kg
hanging mass ( m2 ) = kg
first time interval ( t1 ) =
second time interval ( t2 ) = s
Record t1 on the first row of column 2 and t2 on the first row of column 3 of the Table .
a = 2s ( t1 - t2 ) ⁄ [ t1 t2 ( t1 + t2 ) ]
Using the time intervals, t1 and t2 obtained in procedure (9.) and the distance of one interval ( s ) obtained from procedure (2.), calculate the acceleration of the system.
acceleration ( a ) = 2s ( t1 - t2 ) ⁄ [ t1 t2 ( t1 + t2 ) ] = m ⁄ s 2
Record this on the first row of column 4 of the table.