A flea jumps by exerting a force of 1.33 x 10 N straight down on the ground. A b

Question

A flea jumps by exerting a force of 1.33 x 10 N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 1.17 x 10 N on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is 6.0 x 10 kg. Part 1 of 5 (a) First we need to draw a free-body diagram for the flea as the system and label all forces appropriately. The following questions will help you draw the free-body diagram. We will use m for the mass of the flea, F for the force exerted by the breeze, Fflea for the force the flea exerts on the wind ground, and F for the force the ground exerts on the flea ground For this problem we will assume that the breeze is blowing toward the +x direction. Consider the time interval during which the flea is in contact with the ground. For each force below, if it is a force acting on the flea indicate its direction. (If it is not a force, or if it is a force but does not act on the flea, choose NA.)

Explanation / Answer

a)The net force acting in the vertical direction is:

Fnet-y = F - W

Fnet-y = 1.33 x 10^-5 - 6 x 10^-7 x 9.8 = 7.42 x 10^-6 N

Hence, Fnet-y = 7.42 x 10^-6 N

b)Horizontal force would be:

Fnet-x = 1.17 x 10^-6 N

Resultant force will be:

Fnet = sqrt (Fnet-x^2 + Fnet-y^2)

Fnet = 7.51 x 10^-6 N

theta = tan-1 (Fnet-y/Fnet-x)

theta = tan-1 (7.42 x 10^-6/1.17 x 10^-6) = 81.04 deg

Hence, Fnet = 7.51 x 10^-6 and theta = 81.04 deg

theta = 90 - 81.4 = 8.6 from vertical

For acceleration,

a = F/m = 7.51 x 10^-6/6 x 10^-7 = 12.52 m/s^2

Hence, a = 12.52 m/s^2 ; theta = 8.6 deg

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