Answer:
[tex]F_2 =53.56 \ F_1[/tex]
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 [tex]F_! = ma_1[/tex]
where :
[tex]a_1 = \frac{\delta v_1}{t}[/tex] since F is constant so ; acceleration too is constant
[tex]\delta v_1[/tex] = 2.0 km/h
= [tex]2.0 * \frac{1000 \ m}{3600\ s}[/tex]
= 0.556 m/s
[tex]a_1 = \frac{\delta v_1}{t}[/tex]
[tex]a_1 = \frac{0.5556 }{10}[/tex]
[tex]a_1 =0.0056 \ m/s^2[/tex]
Similarly ;
[tex]F_2 = ma_2[/tex]
We can as well say ;
[tex]a_2 = \frac{\delta v_2}{t}[/tex]
[tex]\delta v_2 = 11 \ km/h[/tex]
[tex]= 11 * \frac{1000 \ m}{3600 \ s}[/tex]
= 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 :
[tex]F_1 + F_2 = ma_2[/tex]
Thus; we can say:
[tex]\frac{F_1 +F_2 }{F_1 } \ = \ \frac{ma_2}{ma_1}[/tex]
[tex]1 + \frac{F_2}{F_1}= \frac{a_2}{a_1}[/tex]
[tex]F_2 = [\frac{a_2}{a_1}-1 ]F_1[/tex]
[tex]F_2 = [ \frac{0.30556 \ m/^2}{0.0056 \ m/s^2} - 1 ] F_1[/tex]
[tex]F_2 = [ \frac{0.30556 \ m/^2 - 0.0056 \ m/s ^2}{0.0056 \ m/s^2}] F_1[/tex]
[tex]F_2 = [ \frac{0.29996 \ m/s ^2}{0.0056 \ m/s^2}] F_1[/tex]
[tex]F_2 =53.56 \ F_1[/tex]
Thus, the magnitude of F1 and F2 compare in a way that says: [tex]F_2 \ is \ aprroximately \ increasing \ at \ the \ rate \ of \ 54 \ times \ of \ F_1[/tex]
