This is coming from a16 year old but I think I'm right "This problem is a Newton's 2nd problem: SF=m*a where SF is the sum of

the external forces acting on the system, m is the total mass

accelerating and a is the rate of acceleration.

In this case the sum of the external forces is equal to the difference

in the weights of the two masses: SF=m1*g-m2*g=(m1-m2)*g while the total

mass accelerating is m1+m2. Therefore, the acceleration of this system

is: A=(m1+m2)/[(m1-m2)*g]" therefore a=(200+300)/[(200/300)*9.8]=

## 76.5306122, so the distance traveled in 5 seconds is

the external forces acting on the system, m is the total mass

accelerating and a is the rate of acceleration.

In this case the sum of the external forces is equal to the difference

in the weights of the two masses: SF=m1*g-m2*g=(m1-m2)*g while the total

mass accelerating is m1+m2. Therefore, the acceleration of this system

is: A=(m1+m2)/[(m1-m2)*g]" therefore a=(200+300)/[(200/300)*9.8]=