A 4.4 kg marble (really big heavy marble) is accelerating down an incline. When it reaches level ground it slows down to a stop in 1.27 seconds due to friction. A. If the deceleration of the marble on level ground was 1.74 m/s/s, how much frictional force was present?

B. Calculate the velocity of the marble when it initially reached level ground. ​

#A

Mass=4.4kg

acceleration=-1.74m/s^2

Use newtons second law

#B

initial velocity=u

Final velocity=v=0

Acceleration=a=-1.74m/s^2

Time=t=1.27s

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