Question

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3 km/h. When a second tugboat applies an additional constant force of magnitude F2 in the same direction, the speed increases by 18 km/h during a 10 s interval. How do the magnitudes of F1 and F2 compare? (Neglect the effects of water resistance and air resistance.)

Answer 1

Answer:

$F_{1}=\frac{1}{5}F_{2}$  or  $F_{2}=5F_{1}$

In other words, $F_{1}$ is one fifth of $F_{2}$ or $F_{2}$ is five times as big as $F_{1}$

Explanation:

In order to solve this problem we must start by sketching the situation (refer to the attached picture).

When the ship is pulled only by force 1, it will change its speed by 3km/hr in 10 seconds. So in order to use these values we need to either turn the km/hr in km/s or turn the seconds to hours. Let's turn the seconds to hours:

$10s*\frac{1hr}{3600s}=\frac{1}{360} hr$

so we can now use the acceleration formula to find the acceleration of the boat so we get:

$a=\frac{\Delta v}{\Delta t}$

which will give us an accceleration of:

$a=\frac{3km/hr}{\frac{1}{360}hr}=1080km/hr^{2}$

once we got the acceleration we can for sure say taht:

$F_{1}=ma=m*1080\frac{km}{hr^{2}}$

Now, if we take a look at the second drawing we can see that the resultant force applied to the boat is found by adding the two forces, force one and force two, so we get:

$F_{1}+F_{2}=ma$

in this case the acceleration changes because the change in velocity is of 18km/hr in the same 10 seconds, so we get that:

$a=\frac{\Delta v}{\Delta t}$

$a=\frac{18km/hr}{\frac{1}{360}hr}=6480km/hr^{2}$

so we can say that:

$F_{1}+F_{2}=m*6480km/hr^{2}$

we can substitute the first force into this equation so we get:

$m*1080km/hr^{2}+F_{2}=m*6480km/hr^{2}$

and solve for the second force, so we get:

$F_{2}=m*6480km/hr^{2}-m*1080km/hr^{2}$

which yields:

$F_{2}=m*5400km/hr^{2}$

Now we can compare theh two forces, force 1 and force 2 by dividing them:

$\frac{F_{1}}{F_{2}}=\frac{m*1080km/hr^{2}}{m*5400km/hr^{2}}$

which yields:

$\frac{F_{1}}{F_{2}}=\frac{1}{5}$

when solving for the first force we get:

$F_{1}=\frac{1}{5}F_{2}$

which tells us that the second force is one fifth of the first force.

and when solving for the second force we get that:

$F_{2}=5F_{1}$

which means that the second force is 5 times as big as the first force.