Question

A flea jumps by exerting a force of Fflea = −1.54 ✕ 10−5 ĵ N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of Fwind = −7.12 ✕ 10−5 î + 2.38 ✕ 10−5 k N on the flea. Find the acceleration of the flea if its mass is 6.00 ✕ 10−7 kg. Express your answer in vector form. Assume that the +x axis is to the right, the +y axis is up along the page, and the +z axis is out of the screen towards you as shown in the figure below

Answer 1

Using Newton's 2nd law:

F_net = m*a

a = acceleration of flea = F_net/m

m = mass of flea = 6.00*10^-7 kg

F_net = F_flea + F_wind

Using given values:

F_net = (-1.54*10^-5 j) + (-7.12*10^-5 i + 2.38*10^-5 k)

F_net = (-7.12 i - 1.54 j + 2.38 k)*10^-5 N

So, acceleration will be:

a = F_net/m

a = (-7.12 i - 1.54 j + 2.38 k)*10^-5/(6.00*10^-7)

a = (-7.12/6.00 i - 1.54/6.00 j + 2.38/6.00 k)*10^2

a = (-1.187 i - 0.2567 j + 0.3967 k)*10^2 m/s^2

a = (-118.7 i - 25.67 j + 39.67 k) m/s^2

In three significant figures

a = (-119 i - 25.7 j + 39.7 k) m/s^2

Let me know if you've any query.