A force in the +x-direction with magnitude F(x) = 18.0\;{\rm N}-(0.530\;{\rm N/m})x is applied to a 8.40kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.

 

If the box is initially at rest at x = 0, what is its speed after it has traveled 16.0m ?

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I understand N to be newtons of force (mass in kg times acceleration in metres per sec per sec). Therefore the force applied for 16m (x=16m) is 18-0.530*16 newtons=9.52 newtons=8.40*acceleration. The acceleration is 9.52/8.40m/sec^2=1.1333m/s^2.

Speed is acceleration times time, so after time t the speed will be 1.1333t. However, the box starts from rest, so to work out the distance travelled we need to take the average speed, which is the average of zero plus 1.1333t metres per second=1/2*1.1333t=0.5667t m/s. We know the distance (=speed times time) travelled is 16m so we can work out t=16/0.5667t seconds, so t^2=28.235 and t=sqrt(28.235)=5.3137s. The speed attained after this amount of time is 5.3137*1.1333=6.0222 m/s.

Another way of working out the speed is to eliminate time as follows. Let's call the constant acceleration a, the speed v, and the distance is x. After time t, v=at, so t=v/a. The average speed is v/2. The distance is x=vt/2=v^2/(2a) and v^2=2ax, so v=sqrt(2ax). We know a=9.52/8.40 and x=16 so v=sqrt(2*9.52*16/8.40)=sqrt(36.2667)=6.0222 metres per second.

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