Five persons A,B,C,D & E are pulling a cart of mass 100Kg on a smooth surface and cart is moving with acceleration 3m/s² in est direction. When person
A stops pulling it moves with acceleration 1m/s² in west direction. When person B stops pulling, it moves with acceleration 24m/s² in the north direction. What is the magnitude of acceleration of the cart when only A&B pull the cart keeping their directions same as the old direction is




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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 Newtons.  --- (1)
when A does not pull the cart,
    vector sum F2 + F3 + F4 + F5 = 100 * 1 * ( - i ) = - 100 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 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 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

2 5 2
Sir the answer is 25m/s2
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