A 10-kg block is being dragged on a horizontal surface with a coefficient of kin

Question

A 10-kg block is being dragged on a horizontal surface with a coefficient of kinetic friction equal to 0.3. The force is 50N, is applied in different ways for a distance of 2 m. a) Determine the speed of the block after the force as been applied for 2 meters if the force is applied horizontally. b) Redo the problem with the force applied (pulled from the right) at an angle of 37 degree with the horizontal. Compare the speed to the answer in (a). c) Is there an angle that maximizes the final kinetic energy of the block? What is it? d) Does the maximizing angle depend on mu_k or on the mass of the block? e) Repeat the problem with the force being used to push the block at an angle of 37 degree. Are there any limits on the angle here?

Explanation / Answer

Q3.

part a:

if the force is applied horizontally, normal force=weight of the block

=10*9.8=98 N

friction force =friction coefficient*normal force

=0.3*98=29.4 N

total force along horizontal direction=applied force - friction force

=50-29.4=20.6 N

acceleration of the block=total force/mass

=20.6/10=2.06 m/s^2

initial speed=0

then final speed=sqrt(initial speed^2+2*acceleration*distance)

=sqrt(2*2.06*2)

=2.8705 m/s

part b:

component of force aong vertical=50*sin(37)=30 N

then normal force=weight-30

=10*9.8-30=68 N

friction force=0.3*68=20.4 N

component of pulling force along horizontal=50*cos(37)=40 N

so total pulling force along horizontal=40-20.4=19.6 N

acceleration=19.6/10=1.96 m/s^2

final speed=sqrt(2*1.96*2)

=2.8 m/s

so the final speed is lesser than part a.

part c:

let angle that maximizes kinetic energy be t degrees.

then friction force=0.3*(10*9.8-50*sin(t))

=29.4-15*sin(t)

total force along the horizontal

F=50*cos(t)-friction force

==>F=50*cos(t)-29.4+15*sin(t)

final kinetic energy will be maximum, if final speed is maximum and final speed is maximum , if applied force is maximum

hence dF/dt=0

==>-50*sin(t)+15*cos(t)=0

==>50*sin(t)=15*cos(t)

==>tan(t)=0.3

==>t=16.7 degrees

so if angle with horizontal is 16.7 degrees, final kinetic energy will be maximum

part d:

maximizing angle depends upon the frction coefficient.

part e:

part e:

normal force=10*9.8+50*sin(37)=128 N

friction force=0.3*128=38.4 N

total force=50*cos(37)-38.4=1.6 N

acceleration=1.6/10=0.16 m/s^2

final speed=sqrt(2*0.16*2)=0.8 m/s

angle should not be such that friction force is greater than the horizontal component of the pushing force

or else the block wont move.

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