When I work that problem out, I get 383645J, but the answer in the book says 386,000J. Am I doing something wrong? If I use 27.7 m/sec in the equation, I get
1/2(1000kg)(27.7m/sec)=383645
How much work is required to stop a 1000 kg car traveling at 100 km/hr?
I was going to use KE=1/2mv^2 , since the work should equal the KE, assuming no friction.
I have: =1/2(1000 kg)(100km/hr)^2
I was using dimensional analysis to make the conversion to m/s, so I end up with J, but...
Then maybe I'm not setting up the conversion correctly, I was using dimensional analysis and set it up
(100 km/hr) (1000m) (3600 sec) = 3.6 x 10^8
(1 km) (1 hr)
but I know this is wrong, by the time you square it and finish the equation, its way too...
How much work is required to stop a 1000 kg car traveling at 100 km/hr?
What I was thinking was that I would use the formula for KE,
KE=1/2mv2 (last two of course is for squared
since the KE would equal work. I have gotten as far as:
KE= 1/2(1000 kg)(100km/hr)2
I am...
I have been working on this for some time, maybe I'm just using the wrong formula.
A 120 kg tackler traveling 3 m/sec tackles a 75 kg halfback running 6m/sec in the opposite direction. What is their common speed immediately after the collison?
Thank you.