Question

A block with mass m = 1.47 kg is placed against a spring on a frictionless incline with angle 0 = 37.10 (see the figure). (The block is not attached to the spring.) The spring, with spring constant k = 19 N/cm, is compressed 22.5 cm and then released. (a) What is the elastic potential energy of the compressed spring? (b) 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? (c) How far along the incline is the highest point from the release point? PA o

Answer 1

Given the mass of the block m = 1.47kg the angle of inclination $\theta = 37.1\degree$ , the spring constant k = 19N/cm = 1900 N/m and the compression of the spring is x = 22.5cm = 22.5 x 10-2m.

(a) The elastic potential energy of the compressed spring is

$PE_{Spring}=\frac{1}{2}kx^{2}$

$PE_{Spring}=\frac{1}{2}\times 1900\times (22.5\times 10^{-2})^{2}=48.09J$

So the elastic pothntial energy of the spring is 48.09J.

(b) Let h1 be the height of the block at the release point (Point 1) and h2 be that at the highest point (Point 2). Applying law of conservation of energy,

Total energy at release point = Total energy at highest point

$KE_{1}+PE_{1}=KE_{2}+PE_{2}$

Since the block is at rest before release and it comes o rest at highest point, the KE at point 1 and 2 is zero. Therefore,

$PE_{1}=PE_{2}$

$(PE_{1})_{Spring}+(PE_{1})_{Block}=(PE_{2})_{Block}$

$(PE_{2})_{Block}-(PE_{1})_{Block}=(PE_{1})_{Spring}$

$(\Delta PE)_{Block}=\frac{1}{2}kx^{2}=48.09J$

So the change in gravitational potentia energy of the block-earth system as the block moves from the release point to its highest point is 4.81mJ.

(c) We know that,

$(\Delta PE)_{Block}=48.09J$

$mgh_{2}-mgh_{1}=48.09$

$mg(h_{2}-h_{1})=48.09$

$1.47\times 9.8\times(h_{2}-h_{1})=48.09$

$14.41\times(h_{2}-h_{1})=48.09$

$(h_{2}-h_{1})=\frac{48.09}{14.41}=3.34m$

So the difference in height between elease point and highest point is 3.34m. Let L be the distance travelled by the block in the incline, then

$\sin \theta=\frac{(h_{2}-h_{1})}{L}$

$L=\frac{(h_{2}-h_{1})}{\sin \theta}=\frac{3.34}{\sin 37.1\degree}=5.54m$

So the block travelled 5.54m up the incline from release point.