Answer to Question #151983 in Classical Mechanics for kerem

Question

A block of mass 2 kg initially compresses a spring (k1 = 400 N/m) by 25 cm as shown below. When the
block is released, it follows a bumpy path until it reaches a second spring. Throughout the path, there is a constant
frictional force of 4 N only on the surface of the hill but the horizontal sections are frictionless. The highest point of
the hill is 24 cm above the ground and the path length from point A to C is 90 cm in total (AB = BC = 45 cm).
a) Find the speed of the object at point B.
b) Find the speed of the object at point C.
c) Find the speed of the object when the 2nd spring (k2 = 250 N/m) is compressed by 15 cm.

Answer 1

As per the given question,

Mass of the block (m)= 2kg

Spring constant $(K_1)=400N/m$

Friction force $(f_s)=4N$

Applying the conservation of energy

a)

$\frac{K_1 x^2}{2}=f_s.d+\frac{mv^2}{2}+mgh$

Now, substituting the values in the equation,

$\Rightarrow \frac{400\times (0.25)^2}{2}=4\times0.45+\frac{2\times v^2}{2}+2\times 9.8\times 0.24$

$\Rightarrow v^2 = 12.5-6.504$

$\Rightarrow v =\sqrt{5.9} m/s$

b) Let the speed of the object at the point C is V

Now, again applying conservation of energy

$\frac{K_1 x^2}{2}=f_s 2d+\frac{mV^2}{2}$

$\Rightarrow 12.5=4\times 0.90+\frac{2V^2}{2}$

$\Rightarrow V^2=12.5-3.6$

$\Rightarrow V=\sqrt{8.9}m/s$

$\Rightarrow V=2.98 m/s$

c) Energy stored in the spring, after the compression

$KE=\frac{K_2 x_2^2}{2}$

$=\frac{250\times 0.15^2}{2}$

$=2.8125J$

Now, applying the conservation of energy

$12.5-4\times 0.9-2.8125 = \frac{2\times V_2^2}{2}$

$\Rightarrow V_2^2=6.0875$

$\Rightarrow V_2=\sqrt{6.0875}m/s$

$\Rightarrow V_2 =2.467 m/s$