IP An 84-kg lumberjack stands at one end of a 380-kg floating log, as shown in t

Question

IP An 84-kg lumberjack stands at one end of a 380-kg floating log, as shown in the figure (Figure!). Both the log and the lumberjack are at rest initially. If the lumberjack now trots toward the other end of the log with a speed of 2.7 m/s relative to the log, what is the lumberjack's speed relative to the shore? Ignore friction between the log and the water. Express your answer using two significant figures. upsilon = m/s If the mass of the log had been greater, would the lumberjack's speed relative to the shore be greater than, less than, or the same as in part A? Greater than in part A Less than in part A The same as in part A Check your answer to part B by calculating the lumberjack's speed relative to the shore for the case of a 500-kg log. Express your answer using two significant figures. Upsilon = m/s

Explanation / Answer

if log goes back with speed v then
speed of lumberjack w.r.t to log = vLj - vlog   = 2.7
vLj   - (-v) = 2.7
v = 2.7 - Vlj
using momentum conservation,
m x vLj = Mv
m x vLj   = M x (2.7 - VLj)
2.7M / (M + m) = vLJ
vLj = 2.7 x 380 / (380 + 84) = 2.21 m/s


B.   2.7M / (M + m) = vLJ   
vLj = 2.7 / (1 + m/M)
if M is greater => m/M is smaller => 1 + m/M will be smaller =? 1 / (m+M) will b greater
so greater than part A .

C. vlj = 2.7 ( 1 + 84 /500) = 2.31 m/s

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