In: Physics
A cookie jar is moving up a 38 ° incline. At a point 51 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.5 m/s. The coefficient of kinetic friction between jar and incline is 0.12. (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?
part a:
mass of the jar=m kg
normal force=m*g*cos(theta) N
friction force=frictino coefficent*normal force
=0.12*m*g*cos(theta)
and it is acting along the downward direction of the incline.
then total force along the incline in downward direction=m*g*sin(theta)+0.12*m*g*cos(theta)
then acceleration in the downward direction=force/mass
=9.8*sin(38)+0.12*9.8*cos(38)=6.96 m/s^2
assuming the downward direction of the inclince to be -ve direction,
initial velocity=u=1.5 m/s
acceleration=-6.96 m/s
final velocity=0 m/s
then distance travelled=(final velocity^2-initial velocity^2)/(2*acceleration)
=0.1616 m
hence it will move 0.1616 m further on the inclince before coming to rest.
part b:
total distance to be travelled to reach the bottom from the top position=0.51+0.1616=0.6716 m
friction force will be acting in the opposite direction of motion i.e. along the upward direction of the inclince.
then total force in downward direction=m*g*sin(theta)-0.12*m*g*cos(theta)
acceleration=force/mass=9.8*sin(38)-0.12*9.8*cos(38)=5.10678 m/s^2
initial velocity=0
distance=0.6716 m
then final velocity=sqrt(initial velocity^2+2*acceleration*distance)=2.619 m/s
hence speed at the bottom is 2.619 m/s
part c:
if coefficient of kinetic friction is decreased, then friction force decreases.
so in first case, (when moving up), the downward acceleration decreases and the jar will move more distance than earlier.
for second case (when moving down), downward acceleration increases and it will have more speed than earlier.