A2.00-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 n/m. the object is displaced 3.00 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) what is the force (magnitude and direction) acting on the object 3.50 s after it is released? (b) how many times does the object oscillate in 3.50 s?

Force = 10.244 Newtons

b) No of oscillations = 0.88

Explanation:

Since the block executes SHM we can write it's position as function of time as

ω is the natural frequency of the system

A is the amplitude of the system

Thus accleration of the block

Thus using the given values at t= 3.50 sec we can calculate the acceleration as

thus force can be calculated using newtons second law as

b)

Now no of oscillations can be obtained as

no of oscillations in 3.50 seconds = 3.50/3.976 = 0.88

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i just did this and got a 100.

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