Astring pulls a 9 kg box horizontally across a table with a force of 50 n. the box has a coefficient of friction with the table of 0.45. what is the net force of the box?

a. 90 n
b. 40 n
c. 50 n
d. 10 n

D. 10 N

Explanation:

Draw a free body diagram of the box.  There are four forces:

Normal force N pushing up.

Weight force mg pulling down.

Applied force P to the right.

Friction force Nμ to the left.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction:

∑F = P − Nμ

∑F = P − mgμ

∑F = (50 N) − (9 kg) (9.8 m/s²) (0.45)

∑F ≈ 10 N

the answer is to check all of them! they all apply well at least i think so!

no. dot product   =0 when vecs at right angles