Can the KE of a car be converted into work by stopping it at 100 km/hr?

AI Thread Summary
To stop a 1000 kg car traveling at 100 km/hr, the kinetic energy (KE) can be calculated using the formula KE = 1/2mv². The correct conversion of speed from km/hr to m/s is essential for accurate calculations, where 100 km/hr converts to approximately 27.78 m/s. After conversion, the kinetic energy should be calculated as KE = 1/2(1000 kg)(27.78 m/s)², resulting in approximately 386,000 J. The discussion highlights the importance of proper unit conversion to avoid errors in calculations. Ultimately, the work done to stop the car equals the kinetic energy, confirming that all KE is converted into work.
rueberry
Messages
5
Reaction score
0
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 assuming that somewhere there is a conversion needed, I'm just not getting the next steps.
 
Physics news on Phys.org
Which next steps? Just convert the km/hr into m/s and you're done.
 
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 big. The answer in the book is 386,000J, but I keep getting answers with 360 b/c of the sec. I'm sure this is a simple mistake I'm making, but I'm still stuck. Sorry.
 
the hour is on the bottom!
your conversion factors should be
(1000 m)/(1 km) * (1 hr)/(3600 s)

I suppose here, 100% of the Work removes KE.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top