I dont understand how to find the work done

[snash1057]

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
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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

 

[SammyS]

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

 

[snash1057]

is it .5 rather?

 

[snash1057]

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!

 

[rwisz]

I can provide a more detailed explanation on how to find the work done in this scenario. Work is defined as the force applied to an object multiplied by the distance it moves in the direction of that force. In this case, the force being applied is the braking force, and the distance it moves is the distance it takes for the car to come to a stop.

To calculate the work done by the brakes, we can use the formula W = Fd, where W is the work, F is the force, and d is the distance. In this scenario, the force is the net force on the car, which is equal to the mass of the car multiplied by its acceleration. We can use the equation F = ma to calculate the force. Since the car is accelerating in the negative direction (due to the braking), the acceleration will be negative.

To find the acceleration, we can use the equation v = u + at, where v is the final velocity (which is 0 in this case), u is the initial velocity (which is the velocity of the car before braking), a is the acceleration, and t is the time it takes for the car to come to a stop. Rearranging this equation to solve for acceleration, we get a = (v-u)/t. Since the car is initially traveling at a constant speed, u will be equal to the final velocity before braking. So, we can simplify the equation to a = -v/t.

Plugging in the given values, we get a = -0/10 = 0 m/s^2. This means that the car is decelerating at a constant rate of 0 m/s^2 due to the braking. Now, we can calculate the force by using the equation F = ma. Since the mass of the car is given as 7475 N, the force will be equal to 7475 N x 0 m/s^2 = 0 N.

Finally, we can use the equation W = Fd to calculate the work done by the brakes. Since the force is 0 N, the work done will also be 0 J. This may seem counterintuitive, but it makes sense if we think about it. The car is coming to a stop, so the brakes are exerting a force to oppose the motion of the car. However, since the car is not moving, there is no displacement and therefore no work is done.

In conclusion