Answer :
2th Law of Newton:
[tex]\vec{R}=m.a\\ \\ \vec{R}=\vec{F} -\vec{F_a}=m.a\\ \\ \vec{F}-15=16.5*2.31\\ \\ \vec{F}=38.115+15\\ \\ \boxed{\vec{F}=53.115 \ N}[/tex]
[tex]\vec{R}=m.a\\ \\ \vec{R}=\vec{F} -\vec{F_a}=m.a\\ \\ \vec{F}-15=16.5*2.31\\ \\ \vec{F}=38.115+15\\ \\ \boxed{\vec{F}=53.115 \ N}[/tex]
Call the applied force 'A'. (Clever ?)
The forces on the cart are 'A' forward and 15 N backward.
The net force on the art is (A-15) forward.
F = m a
Net forward force = (mass of the cart) x (its forward acceleration)
(A - 15) = (16.5) x (2.31)
A - 15 = 38.115
Add 15 to each side:
A = 53.115 newtons
The forces on the cart are 'A' forward and 15 N backward.
The net force on the art is (A-15) forward.
F = m a
Net forward force = (mass of the cart) x (its forward acceleration)
(A - 15) = (16.5) x (2.31)
A - 15 = 38.115
Add 15 to each side:
A = 53.115 newtons
