Question

An ice cube of mass 50.0g can slide without friction up and down a 25.0 degree slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 0.100m . The spring constant is 25.0 N/m . When the ice cube is released, how far will it travel up the slope before reversing direction?
A. Identify the initial and final gravitational potential energies.
B. Identify the initial and final elastic potential energies.

Answer 1

Answer:

Explanation:

Given

mass of ice cube $m=50 gm$

inclination $\theta =25^{\circ}$

Spring constant $k=25\ N/m$

initial compression $x=0.1 m$

Suppose Initially the cube is at the datum  and it moves h height after release

So Elastic Potential Energy of Spring will convert to Potential Energy of cube

$\frac{1}{2}kx^2=mgh$

$\frac{1}{2}\times 25\times (0.1)^2=50\times 10^{-3}\times 9.8\times h$

$h=0.255 m$

so final Potential Energy will be $=mgh=0.05\times 9.8\times 0.255=0.125 J$

Initial Potential Energy$=0 J$

B.Initially spring is at compression so energy associated with it is its Elastic Potential Energy i.e. 0.125 J

Final Energy$=0 J$  

Answer 2

Answer: distance = 0.6 m

Explanation:

Given that:

Mass = 50g = 0.05kg

Ø = 25 degree

K = 25 N/m

Extension x = 0.1m

Let us first consider the total energy in the system.

Sum of the total energy = 0

That is

Final E - Initial E = 0

K.E, P.E and energy in the spring.

This equation will be reduced to

Mgh - 1/2mv^2 - 1/2Kx^2 = 0

Final velocity = 0 because of the spring energy

Initial height = 0

Final spring E = 0 as the spring returns to it normal length. Therefore,

Mgh - 1/2mv^2 - 1/2Kx^2 = 0

Since the ice cube was initially at rest, initial V = 0

Mgh = 1/2Kx^2

0.05 × 9.81 × h = 0.5 × 25 × 0.1^2

h = 0.125/0.4905

h = 0.255 m

how far will it travel up the slope before reversing direction?

Using SohCahToa for a right angle

SinØ = h/d

Where d = distance travel

Sin25 = 0.255/d

d = 0.255/sin 25

d = 0.603 m