Question

An ice cube of mass 50.0 g 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 0.100 m 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?
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the spring to a distance of 0.200 m, how far up the slope will an identical ice cube travel before reversing directions?

Answer 1

Answer:

a) $s = 0.603\,m$, b) $s = 2.412\,m$

Explanation:

a) The system ice cube-spring is modelled by means of the Principle of Energy Conservation. Let assume that height at the bottom is zero:

$U_{g} = U_{k}$

$m\cdot g \cdot s\cdot \sin \theta = \frac{1}{2}\cdot k \cdot x^{2}$

$s = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \sin \theta}$

$s = \frac{(25\,\frac{N}{m} )\cdot (0.1\,m)^{2}}{2\cdot (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 25^{\textdegree}}$

$s = 0.603\,m$

b) The distance travelled by the ice cube is:

$s = \frac{(25\,\frac{N}{m} )\cdot (0.2\,m)^{2}}{2\cdot (0.05\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 25^{\textdegree}}$

$s = 2.412\,m$

Answer 2

Answer:

A) at x =0.1m, distance it will travel up the slope before reversing direction = 0.603m

B) at x =0.2m, distance it will travel up the slope before reversing direction = 2.412m

Explanation:

A) First of all, the formula for Elastic potential energy is given as;

P.E = (1/2)kx²

where:

P.E= elastic potential energy ( J )

k = spring constant ( N/m )

x = spring extension or compression (m)

From the question, we are given;

mass of ice cube (m) = 50g = 0.05 kg

angle of ramp slope (θ) = 25.0°

compression of the spring (x) = 0.1m

spring constant (k =) 25.0 N/m

From conservation of energy, we know that;

Elastic Potential Energy = Gravitational Potential Energy

Thus;

(1/2)kx² = mgh

From the free body diagram i attached, it is clear that h/d = sin 25°

h = d sin25

Thus, plugging in the relevant values, we have;

(1/2)(25)(0.1)² = 0.05 x 9.81 x d x sin 25

0.125 = 0.4905 x 0.4226 x d

d = 0.125/(0.4905 x 0.4226) = 0.603m

B) if the spring is now compressed to a distance of 0.2m, x = 0.2m

Thus; (1/2)kx² = mgh will now be;

(1/2)(25)(0.2)² = 0.05 x 9.81 x d x sin 25

0.5 = 0.4905 x 0.4226 x d

d = 0.5/(0.4905 x 0.4226) = 2.412m