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 2.0 km/h. When a second tugboat applies an additional constant force of magnitude F2 in the same direction, the speed increases by 11 km/h during a 10 s interval. How do the magnitudes of F1 and F2 compare

Answer 1

`F_2 =53.56 \ F_1`

Step-by-step explanation:

Given that:

A tugboat tows a ship with a constant force of magnitude F1 and then increases the ship's speed during a 10 s interval with 2.0 km/h.

If F = ma

we can say for
`F_! = ma_1`

where :

`a_1 = (\delta v_1)/(t)` since F is constant so ; acceleration too is constant

`\delta v_1` = 2.0 km/h

=
`2.0 * (1000 \ m)/(3600\ s)`

= 0.556 m/s

`a_1 = (\delta v_1)/(t)`

`a_1 = (0.5556 )/(10)`

`a_1 =0.0056 \ m/s^2`

Similarly ;

`F_2 = ma_2`

We can as well say ;

`a_2 = (\delta v_2)/(t)`

`\delta v_2 = 11 \ km/h`

`= 11 * (1000 \ m)/(3600 \ s)`

= 3.0556 m/s²

Now; the additional force acting on the ship in the same direction of the force i.e the net force is :

`F_1 + F_2 = ma_2`

Thus; we can say:

`(F_1 +F_2 )/(F_1 ) \ = \ (ma_2)/(ma_1)`

`1 + (F_2)/(F_1)= (a_2)/(a_1)`

`F_2 = [(a_2)/(a_1)-1 ]F_1`

`F_2 = [ (0.30556 \ m/^2)/(0.0056 \ m/s^2) - 1 ] F_1`

`F_2 = [ (0.30556 \ m/^2 - 0.0056 \ m/s ^2)/(0.0056 \ m/s^2)] F_1`

`F_2 = [ (0.29996 \ m/s ^2)/(0.0056 \ m/s^2)] F_1`

`F_2 =53.56 \ F_1`

Thus, the magnitude of F1 and F2 compare in a way that says:
`F_2 \ is \ aprroximately \ increasing \ at \ the \ rate \ of \ 54 \ times \ of \ F_1`