I dont understand how to find the work done

  • Thread starter Thread starter snash1057
  • Start date Start date
  • Tags Tags
    Work Work done
AI Thread Summary
To find the work done by the brakes on a car, the driver needs to calculate the correct acceleration using the formula for distance and time. The initial attempt to find acceleration was incorrect, leading to mistakes in calculating force and work. After correcting the algebra, the driver successfully determined the right values, leading to the correct answer. Understanding the relationship between acceleration, force, and work is crucial in solving such physics problems. Accurate calculations are essential for determining the work done in stopping the vehicle.
snash1057
Messages
15
Reaction score
0
i don't understand how to find the "work" done

A driver of a 7475 N car passes a sign stating "Bridge Out 25 Meters Ahead." She slams on the brakes, coming to a stop in 10 s. How much work must be done by the brakes on the car if it is to stop just in time? Neglect the weight of the driver, and assume that the negative acceleration of the car caused by the braking is constant.W=Fdcos
i tried finding acceleration by doing 25=1/2a(10^2) = .125
then finding mass by doing 7475=m(9.8)= 762.76
and then Fnet = 762.76 x .125 = 95.35
w= 95.35(25) = 2383.63
___________________________________________________

i also tried finding Vf by using 25=1/2(0+Vf)10 = 1.25

then i tried finding kinetic energy by 1/2(7475)(1.25^2)= 5839.84but neither of them are right so i have to be doing something wrong
 
Last edited:
Physics news on Phys.org


You had 25=1/2a(10^2) = .125 . ?I think you meant that a = .125, but that's wrong. Check your algebra.
 


is it .5 rather?
 


OH GOSH! i can't believe i made that dumb of a mistake! haha i got the right answer! thank you for correcting my algebra!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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