Question

A 0.17- kg block on a wooden table is given a sharp blow with a hammer. The block then slides across the table to a stop. The coefficient of friction between the block and table is 0.38. The hammer's force on the block as a function of time is well approximated by an isosceles triangle with it's base on the horizontal time axis. The length of the base is 0.0074 s and the maximum force is 48.5 N. How far does the block go? Use a graphical technique.

Answer 1

impulse applied=area of force-time curve

=area of the triange

=0.5*base*height=0.5*0.0074*48.5=0.17945 kg.m/s

as we know, impulse applied=change in momentum

as initial speed is zero,

if final speed is v,

then mass*v=impulse applied

0.17*v=0.17945

==>v=0.17945/0.17=1.0556 m/s

hence when the block starts moving, its speed is 1.0556 m/s

then its kinetic energy=0.5*mass*v^2

=0.5*0.17*1.0556^2=0.094715 J

this energy is utilized as work done against friction.

hence if the block travels a distance d before coming to rest,

work done against friction for travelling distance d=initial kinetic energy of the block

friction force=kinetic friction coefficient*normal force

normal force=weight of the block=0.17*9.8=1.666 N

then friction force=0.38*1.666=0.6331 N

hence work done against friction for travelling d m=0.6331*d J

hence 0.6331*d=initial kinetic energy=0.094715

==>d=0.094715/0.6331=0.14961 m=14.961 cm

hence the block travels 14.961 cm before coming to rest.