A 2.00 kg block is placed against a spring on a friction-less 24 degree incline. The spring,...

Question:

A 2.00 kg block is placed against a spring on a friction-less 24{eq}^{\circ} {/eq} incline. The spring, whose spring constant is 19.8 N/cm, is compressed 24.6 cm and then released.

What is the elastic potential energy of the compressed spring?

What is the change in the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline?

How far along the incline is the highest point from the release point?

Energy conservation

We know that from the law of energy conservation that energy can neither be created nor be destroyed it can only change its forms from one to another. Therefore, the total energy of the system would remain constant at every instant.

Given

Mass of the block (m) = 2 kg

Spring Constant (k) = 1980 N/m

Spring compression (x) = 24.6 cm

(a)

Spring potential energy

{eq}\displaystyle E = 0.5kx^{2} \\ E = 0.5*1980*(0.246)^{2} = 59.91 \ J {/eq}

(b)

Now, from energy conservation, the gravitational potential energy of the block will be

{eq}\displaystyle U_{b} = E \\ U_{b} = 59.91 \ J {/eq}

(c)

Now we know that the gravitational potential energy is given by

{eq}\displaystyle U_{b} = mgh \\ 59.91 = 2*9.81*h \\ h = 3.05 \ m {/eq}

Now the length along incline would be

{eq}\displaystyle L\sin\theta = h \\ L\sin24 = 3.05 \\ L = 7.51 \ m {/eq}