Aspring is mounted horizontally on an air track, with the left end held stationary. we attach a spring balance to the free end of the spring, pull toward the right, and measure the elongation. we determine that the stretching force is proportional to the displacement and that a force of 6.0 n causes an elongation of 0.030 m. we remove the spring balance and attach a 0.50 kg object to the end, pull it a distance of 0.040 m, release it, and watch it oscillate in shm. find the following quantities: a. the force constant of the springb. the maximum and minimum velocities attained by the vibrating objectc. the maximum and minimum accelerationsd. the velocity and acceleration when the object has moved halfway to the center from its initial positione. the kinetic energy, potential energy, and total energy in the halfway position

Explanation:

a ) A force of 6 N , causes elongation of .03 m

Spring constant

k = 6 / .03

= 200 N / m

b )

Amplitude A = .04 m .

Maximum velocity of a particle in SHM

= ωA where ω is angular velocity and A is amplitude

= x A

= x A

= 20x .04 = 0.8

Maximum velocity of a particle

= 0.8 m ² s

Minimum velocity will be zero at extreme point ( at turning point )

c )

Minimum acceleration is zero at middle point ( equilibrium position )

maximum acceleration

= ω²A

=  k / m x A

= 400 x .04

= 16 m / s

d )

At halfway point velocity will be √ 3 / 2 times the maximum velocity

velocity at mid point

= √ 3 / 2 x max velocity

.866 x

= .866 x  0.8

= .693 m /s

the answer is neither ball has a charge.

because if they were both positively charged, they will repel each other.

if they were positively and negatively charged, they will seek.

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