Question

008 (part 1 of 3) 10.0 points A 3.2 kg object is subjected to two forces, F~ 1 = (1.9 N) ˆı + (−1.9 N) ˆ and F~ 2 = (3.8 N) ˆı + (−10.1 N) ˆ. The object is at rest at the origin at time t = 0. What is the magnitude of the object’s acceleration? Answer in units of m/s 2 .

Answer 1

a' =4.15 m/s²

Step-by-step explanation:

Given that

m= 3.2 kg

F₁ = 1.9 i −1.9 j N

F₂=3.8 i −10.1 j N

From second law of Newton's

F(net) = m a

F₁ + F₂ = m x a

1.9 i −1.9 j + 3.8 i −10.1 j = 3.2 a

a = 1.78 i - 3.75 j m/s²

The resultant acceleration a'

`a'=√(1.78^2+3.75^2)\ m/s^2`

a' =4.15 m/s²