In the figure here, a box of Cheerios (mass mC = 1.6 kg) and a box of Wheaties (
ID: 2219901 • Letter: I
Question
In the figure here, a box of Cheerios (mass mC = 1.6 kg) and a box of Wheaties (mass mW = 2.7 kg) are accelerated across a horizontal surface by a horizontal force applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 2.6 N, and the magnitude of the frictional force on the Wheaties box is 5.1 N. If the magnitude of F is 12.3 N, what is the magnitude of the force on the Wheaties box from the Cheerios box?
In the figure here, a box of Cheerios (mass mC = 1.6 kg) and a box of Wheaties (mass mW = 2.7 kg) are accelerated across a horizontal surface by a horizontal force applied to the Cheerios box. The magnitude of the frictional force on the Cheerios box is 2.6 N, and the magnitude of the frictional force on the Wheaties box is 5.1 N. If the magnitude of F is 12.3 N, what is the magnitude of the force on the Wheaties box from the Cheerios box?Explanation / Answer
By F(net) = F(applied) - F(friction)
=>(mc+mw) x a = 12 - (2+3.5)
=>(1.6 + 3.3) x a = 6.5
=>a = 1.32 m/s^2
For Wheaties box :-
=>F(net) = F(applied) - F(friction)
=>m x a = F(force by Cheerios box) - 3.5
=>3.3 x 1.32 = F(force by Cheerios box) - 3.5
=>F(force by Cheerios box) = 7.87 Newton
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