Question

The Rafflesia arnoldii found in Malaysia, is the largest flower in the world, one meter across and with a mass of 11 kg. If the coefficient of static friction between the flower and the forest floor is 0.76, find the minimum force you would need to apply to set the flower in motion. If you continue to push the flower with the same force calculated above, and if the coefficient of kinetic friction between the flower and the forest is 0.44, calculate the acceleration of the flower. How far does the flower move in 3.75 seconds?

Answer 1

Answer:

Explanation:

Minimum force needed to set the flower in motion = maximum frictional force

possible on the flower by the floor .

= μ x mg

where μ is coefficient of friction of the floor and mg is weight of the flower

= .76 x 11 x 9.8

= 81.9 N

When the flower is in motion , kinetic friction is applied on it

Force of kinetic friction applied on it

= μk x mg where μk is coefficient of  kinetic friction of the floor

= .44 x 11 x 9.8

= 47.43 N

Pushing force is 81.9 N , so net force on flower when it is in motion

= 81.9 - 47.43 N

= 34.47 N

Acceleration = net force / mass

= 34.47 / 11

= 3.133 m /s

distance moved by flower in 3.75 s

s = ut + 1/2 a t²

= 0 + .5 x 3.133 x 3.75²

= 22 m