A 2.8kg box is filled with 12.8kg of nuts and bolts and slid across the floor with a force of 102N. If the coefficient of friction between the floor and the box is 0.335,
what is the acceleration of the box?

3.26 m/s²

Explanation:

Draw a free body diagram.  There are four forces:

Weight force mg pulling down,

Normal force N pushing up,

Applied force F pushing right,

Friction force Nμ pushing left.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction:

∑F = ma

F − Nμ = ma

F − mgμ = ma

Plug in values:

102 N − (2.8 kg + 12.8 kg) (9.8 m/s²) (0.335) = (2.8 kg + 12.8 kg) a

a = 3.26 m/s²

atomic = no of protons.

mass number-atomic number = neutron number