As a prank, your friends have kidnapped you in your sleep, and transported you out onto the ice covering a local pond. Since you're an engineer, the first thing you do when you wake up is drill a small hole in the ice and estimate the ice to be 6.7cm thick and the distance to the closest shore to be 31.7 m. The ice is so slippery (i.e. frictionless) that you cannot seem to get yourself moving. You realize that you can use Newton's third law to your advantage, and choose to throw the heaviest thing you have, one boot, in order to get yourself moving. Take your weight to be 634 N. (Lucky for you that, as an engineer, you sleep with your knife in your pocket and your boots on.)a. If you throw your 1.32-kg boot with an average force of 457 N, and the throw takes 0.646 s (the time interval over which you apply the force), what is the magnitude of the force that the boot exerts on you? (Assume constant acceleration.)b. How long does it take you to reach shore, including the short time in which you were throwing the boot?

a) As the force of man on the boot is the action and the reaction of the boot on the man is the force of 457 N directed in the opposite direction of the movement of the boot

b) For this part we divide the calculation: a first part when man is accelerated by the reaction force and a second part when this force is zero.

Let's calculate the average acceleration with Newton's second law on man

F = ma

a = F / M

a = 457 / 64.7

a = 7 m/s²

Distance traveled with this acceleration

X1 = Vo t + ½ to t2

The man leaves the rest Vo = 0

X1 = ½ 7 0.646²

X1 = 1.46 m

After the force disappears, we have the speed reached by the impulse

I = f t

I = 457 0.646

I = 295.2 N s

I = MV -M Vo

As man leaves rest Vo = 0

V = I / M

V = 295.2 / 64.7

V = 4.56 m / s

The spped is constant

V = x2 / t2

t2 = x2 / v2

The distance left to travel is

X2 = 31.7 - 1.46

X2 = 30.24 m

t = 30.24 /4.56

t = 6.63s

Finally, the total time is the sum of the two times

t all = 6.63 +0.6

t all = 7.2 s