Originally Posted by **Morgan82** A lobster is in a tug of war against some smaller lobsters.
Its mass = 20.0 kg
The small lobsters drag the big lobster 1.55 m at 0.550 m/s.
Lobster is initally at rest. What is the magnitude of the force applied to the lobster? Ignore friction and water. |

Lets assume constant force and hence acceleration, that the average speed over the $\displaystyle 1.55 $m is $\displaystyle 0.550 $m/s, and the initial speed is $\displaystyle 0$, and the initial position is $\displaystyle s=0$. Also assume movement is in the positive direction so our speed and velocity are the same as long as we are moving to the right.

Then the equation of motions give:

$\displaystyle

v=at

$

$\displaystyle

s=\frac{at^2}{2}

$

and if $\displaystyle t_1$ is the time over which the displacement takes place, the average speed (velocity) is:

$\displaystyle

\bar{v}=0.55=\frac{at_1}{2}

$

and

$\displaystyle

s_1=1.55=\frac{at_1^2}{2}

$

Now solve these for $\displaystyle t_1$ and a, and then use $\displaystyle F=ma$ to find the force.

CB