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If a grocery cart with a mass of 16.5 kg accelerates at +2.31 m/s2 against a frictional force of -15.0 N, what is the applied force?

Sagot :

Ryan2idn
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]
AL2006idn
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

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