Question

If the coefficient of kinetic friction is 0.2, what friction of the total mechanical energy is lost due to frictional forces?

Run#	Mass of hanger + weights (M)	mass of glider +flag (m)	velocity of glider (v)	d (m)	potential energy lost (Mgd) (J)	Kinetic energy gained 1/2 (m+M)v2	% change (ME)
1	0.05	0.2118	0.88	0.235	0.1152675	0.10136896	12.05763984
2	0.07	0.2118	1.01	0.235	0.1613745	0.14373209	10.93258848
3	0.09	0.2118	1.12	0.235	0.2074815	0.18928896	8.768270906
4	0.11	0.2118	1.2	0.235	0.2535885	0.231696	8.633080759
5	0.13	0.2118	1.26	0.235	0.2996955	0.27132084	9.467829847
6	0.15	0.2118	1.34	0.235	0.3458025	0.32482404	6.066601601

Answer 1

Frictional force is due to the weight of the glider . frictional force is given by F =mg

=0.2

For Run #1) Frictional force = mg = 0.2 (0.2118) (9.8) = 0.415128 N

Energy lost due to frictional force = mg *d = 0.415128 ( 0.235) = 0.09755508 j

   Total mechanical energy = potential energy = Mgd + 0.5MU^2 = Mgd + 0 as initial velocity of M is zero .

ME = 0.1152675 J

   Energy lost by friction/ Total mechanical energy = 0.09755508/0.1152675 = 0.85

For Run #2 ) ME = 0.1613745 J

Fraction = Friction energy/ ME = 0.09755508/0.1613745 =0.6

For Run #3 ) ME = 0.2074815 J

Fraction = Friction energy/ ME = 0.09755508/0.2074815=0.47

For Run # 4 ) ME = 0.2535885 J

Fraction = Friction energy/ ME = 0.09755508/0.2532885 =0.385

For Run #5 ) ME = 0.2996955J

Fraction = Friction energy/ ME = 0.09755508/0.2996955 =0.325

For Run #6 ) ME = 0.3458025 J

Fraction = Friction energy/ ME = 0.09755508/0.3458025 =0.28