contestada

You (50 kg) are standing on a floating log (200 kg). Both are floating down a river at 1 m/s. The log points in the direction along the river. You walk in a direction that is down river for 5 seconds and your speed is 1.5 m/s as measured by a ground observer at the end of the 5 seconds.
a. What is the speed of the log at the end of the 5 seconds?
b. What is the average force between you and the log during those 5 seconds?

Respuesta :

boffeemadrid

Answer:

0.875 m/s

5 N

Explanation:

[tex]m_1[/tex] = Mass of person = 50 kg

[tex]m_2[/tex] = Mass of log = 200 kg

[tex]v_1[/tex] = Velocity of person = 1.5 m/s

[tex]v_2[/tex] = Velocity of log

v = Velocity of log with respect to shore = 1 m/s

t = Time taken = 5 seconds

As the momentum of system is conserved we have

[tex](m_1+m_2)v=m_1v_1+m_2v_2\\\Rightarrow v_2=\frac{(m_1+m_2)v-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{(50+200)1-50\times 1.5}{200}\\\Rightarrow v_2=0.875\ m/s[/tex]

Velocity of the log at the end of the 5 seconds is 0.875 m/s

Force is given by

[tex]F=\frac{m(v-u)}{t}\\\Rightarrow F=\frac{50(1.5-1)}{5}\\\Rightarrow F=5\ N[/tex]

The average force between you and the log during those 5 seconds is 5 N