Ablock of mass 0.242 kg is placed on top of a light, vertical spring of force constant 5 050 n/m and pushed downward so that the spring is compressed by 0.091 m. after the block is released from rest, it travels upward and then leaves the spring. to what maximum height above the point of release does it rise?

8.82 m

Explanation:

When compressed, the spring potential energy is:

where k = 5050 N/m is the spring constant, x = 0.091 is the distance compressed

This energy would be converted to kinetic, so the mass gains speed, which is then converted to gravitational potential energy once the mass reaches its highest point, which is 0 speed/kinetic energy

where m = 0.242 kg is the mass of the block, g = 9.8m/s2 is the gravitational acceleration, and h is the maximum height which we are looking for

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