Let F1, F2, F3, F4, and F5 be the forces exerted by the five persons A, B, C, D and E respectively on the cart to pull the cart. These are all vector quantities.
Let the unit vector (force in Newtons) in the direction of East be denoted by i. Let the unit vector in North direction be denoted by j. Then:
when all of them pull the cart:
vector Sum F1 + F2 + F3 + F4 + F5 = 100 * 3 * i = 300 i Newtons. --- (1)
when A does not pull the cart,
vector sum F2 + F3 + F4 + F5 = 100 * 1 * ( - i ) = - 100 i Newtons. --- (2)
Subtract the two above equations from one another: F1 = 400 i .
It is not clear when person B stops pulling, Whether A is pulling the cart or not.. Let us assume that A , C, D and E are pulling the cart with a = 24 m/s/s in North.
vector sum F1 + F3 + F4 + F5 = 100 * 24 * j = 2400 j Newtons --- (3)
eq (1) - (3) gives : F2 = 300 i - 2,400 j Newtons.
If only A and B pull the cart, then the net force on the cart is :
vector sum F1 + F2 = 700 i - 2, 400 j Newtons
The acceleration of the cart will then be : a = F / mass = 7 i - 24 j m/sec²
magnitude of acceleration = √(7² + 24²) = 25 m/sec²