Question

Help please!!! (:
A net force of 2.6N(i-hat) - 2.5N(j-hat) acts on a 0.9 kg object. Find the acceleration a.
=>_______m/s^2(i-hat) + ________m/s^2(j-hat)

Answer 1

\[F - F \mu = ma\]
\[2.6N - 2.1N = 0.9kg(a)\]
\[0.5 = 0.9a\]
\[a = \frac{ 0.5 }{ 0.9 }\]
\[a = 0.556 m/s ^{2}\]

Answer 2

you have \[\vec F = m\vec a\\\vec a=\frac{\vec F}{m}=\frac{(2.6 \hat{i} -2.5\hat{j}}{0.9}= 2.9\hat{i}-2.8\hat{j} (m/s^2)\]
that means to horizontal direction, the acceleration is 2.9 m/s^2 and to the vertical direction, the acceleration is 2.8 m/s^2