Answer to Question #191410 in Mechanics | Relativity for Joseph Belles

Question

The most powerful tugboats in the world are built in Finland. These boats exert a force with a  magnitude of 2.85 x 106 N. Suppose one of these tugboats is trying to slow a huge barge that has  a mass of 2.0 x 107 kg and is moving with a speed of 3.0 m/s. If the tugboat exerts its maximum  

force for 21 s in the direction opposite to that in which the barge is moving, what will be the  change in the barge’s momentum? How far will the barge travel before it is brought to a stop?  

Answer 1

From newton's second law of motion , which states that the rate of change in momentum is directly proportional to the force applied:

$p = F \times t$

$p = -2.85 \times10^{6} \times 21$

$p = -5.985 \times 10^7kg(ms^{-1})$

Since we need to find the final velocity, (v₂), we need to isolate it by solving the equation for this value.

$m(v_2-v_1) = F \Delta t$

$v_2 = \dfrac{F \Delta t}{m} + v_1$

$v_2 = \dfrac{-2.85 \times 10^6 \times 21}{2 \times 10^7}+ 3$

$v_2 = 0.0075ms^{-1}$

Now, using the equation:

$\Delta p = m(v_2-v_1)$

$\Delta p = 2 \times 10^7(0.0075-3)$

$\Delta p = -5.985 \times 10^7 kgms^{-1}$

Now,

$\Delta x = \dfrac{(v_1+v_2)\Delta t}{2}$

$\Delta x = \dfrac{(0.0075+3)21}{2}$

$\Delta x = 3.157m$