A205-g block is pressed against a spring of force constant 1.18 kn/m until the block compresses the spring 10.0 cm. the spring rests at the bottom of a ramp inclined at 60.0° to the horizontal. using energy considerations, determine how far up the incline the block moves from its initial position before it stops under the following conditions

L = 3.391m

Explanation:

a) Given:

Mass of the block, m = 205g=0.205kg

spring constant, k = 1.18 kN/m

Angle made by the ramp = 60°

displacement of the spring, x = 10cm= 0.1m

Now,

The initial potential energy of the spring = final gravitational potential energy of block

mathematically,

where,

g = acceleration due to gravity

h = height of the block above the ground = LsinΘ

L is the displacement of the block parallel to the ramp

substituting the values in the above equation we get

or

or

L = 3.391m

thermal energy is the answer

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