Aspring with a spring constant value of 2500 is compressed 32 cm. a 1.5-kg rock is placed on top of it, then the spring is released. approximately how high will the rock rise? 9 m 17 m 27 m 85 m

9 m

Explanation:

The elastic potential energy initially stored in the spring is given by:

where

k = 2500 N/m is the spring constant

x = 32 cm = 0.32 m is the compression of the spring

Substituting:

Due to the law of conservation of energy, when the spring is released all this energy is converted into kinetic energy of the rock, which starts moving upward. As the rock reaches its maximum height, all the energy has been converted into gravitational potential energy:

where

m = 1.5 kg is the rock's mass

g = 9.8 m/s^2 is the gravitational acceleration

h = ? is the maximum height reached by the rock

Using U=128 J, we find h:

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask

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