A 2150 kg truck traveling north at 42 km/h turns east and accelerates to 61 km/h. (a) What is the change in the truck's kinetic energy? (b) What is the magnitude of the change in its momentum ?

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Answer 1 · 2015-10-10T18:33:50

$Delta E_k=281,2676kJ$
$|Delta vec p|=44230,367kg.m//s$

$DeltaE_k=1/2mv^2-1/2m u^2=1/2m(v^2-u^2)$

$=1/2xx2150xx(16,944^2-11,667^2)$

$=281,2676kJ$

$Delta (vec p)=mvecv-m vecu=m(vecv-vecu)$

#vecp_i=m vec u = 2150 (0 hati+11,667 hat j)=0 hati + 25084,05 hat j kg.m//s#

#vecp_f=m vec v=2150(16,944 hat i + 0 hat j)=36429,6 hati+0 hatj kg.m//s#

$therefore Delta vec p=vecp_f-vecp_i=(36429,6-0)hati + (0-25084,05)hatj$

#=36429,6 hati -25084,05 hatj kg.m//s#

Therefore in magnitude,

$|Delta vecp|=sqrt(36429,6^2+25084,05^2)=44230,367kg.m//s$