A lumberjack with a mass of 105 kg is standing at the left end of a floating log

Question

A lumberjack with a mass of 105 kg is standing at the left end of a floating log which has a mass of 265 kg and is also at rest. The log is 12.0 meters long.

The lumberjack runs to the other end of the log attaining a speed of 3.40 m/s relative tothe shore and then hops on to an identical floating log that is initially at rest. You can ignore any friction or resistance between the logs and the water.

a) Treating the lumberjack as a point mass, locate the centre of mass of the lumberjack-log combination relative to the left end of the log

b) What is the velocity (magnitude and direction) of the first log relative to the shore just before the lumberjack jumps off?

c) What is the velocity (magnitude and direction) of the second log relative to the shore if the lumberjack comes to rest on it?

d) What is the magnitude of the average force exerted by the second log on the lumberjack if he comes to rest relative to the log in 0.280 seconds?

Explanation / Answer

a)Center of mass of Log is at its mid point.So 6m from left end of log

center of mass X=(m1x1+m2x2)/(m1+m2 )

                           =(105*0+265*6)/(105+265)

                           =4.297 meter from left end of log

b)Using conservation of momentum

m1v1+m2v2=0    because initially system is at resy with respect to ground

=>v2=-105*3.4/265=1.35 m/s opposite to velocity of man

c)Using conservation of momentum

105*3.4=V*(105+265)

=>V=105*3.4/(105+265)=0.965 m/s

d)average descelaration=(3.4-0.956)/0.280

So average force =105* (3.4-0.956)/0.280=916.5 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.