A flea jumps by exerting a force of 1.07 10-5 N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 1.14 10-6 N on the flea. Find the direction and magnitude (in m/s2) of the acceleration of the flea if its mass is 6.0 10-7 kg. (Let us assume that).

a) 15.77 m/sec2

b) 13.3 deg

Explanation:

we are given;

Flea force = F1=1.07×10⁻5 N j

Breeze force = F2 = 1.14× 10⁻6 N (-j

mass of flea =6.0 ×10⁻7 kg

So net force on the flea=F1+F2+weight of flea=1.07×10⁻5 j +1.14× 10⁻6 i + 6.0 ×10⁻7 (-j) ×9.8= ma

==> ma = 1.07×10⁻5 j - 0.588×10⁻5 j + 0.114×10⁻5 i

==> ma= 0.482 ×10⁻5 j +0.114×10⁻5 i

==> ma = 0.114×10⁻5 i +0.482 ×10⁻5 j

== a = (0.114×10⁻5 i +0.482 ×10⁻5 j) / 6.0 ×10⁻7

==> a =

==>a= (1.9 j+8.03 i ) m/sec2

mag of a = 15.77 m/sec2

direction angle = tan⁻1(1.9/8.03)=13.3°

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s