(a) Let #l# be length of the inclined plane of #angle theta#.
Height of the inclined plane #h=lsin theta=5.0xxsin37^@#
Potential energy of the mass at the top of inclined plane #PE=mgh=3.0xx9.8xx(5.0xxsin37^@)#
#=>PE=88.5J#
Kinetic energy at the bottom of plane #KE=1/2mv^2#
#=>KE=1/2xx3.0xx(2.0)^2=6.0J#
Loss of energy due to friction #PE-KE=88.5-6.0=82.5J#
If #f# is average force due to friction work done against friction
#W=vecfcdot vecs#
Inserting various values we get
#82.5=|vecf|xx5.0#
#f=82.5/5.0=16.5N#, rounded to one decimal place
(b)
Force of friction #f=muN#
where #mu# is coefficient of friction and #N# is normal reaction.
As shown in the figure #N=mgcostheta#.
From above two equations we get
#mu=f/(mgcostheta)#
Inserting various values we get
#mu=16.5/(3.0xx9.8xxcos37^@)#
#mu=0.7#, rounded to one decimal place.
