PROBLEM 1 - A 0.15-kg frame, when suspended at rest from a coil spring of neglig

Question

PROBLEM 1 - A 0.15-kg frame, when suspended at rest from a coil spring of negligible mass, stretches this spring 0.040 m. A 0.20-kg lump of putty is dropped from rest onto the frame from a height of 30 cm. We define the following four situations: lump # 0.20 kg ? Lump is released frame m, 0.15 kg Lump is about to hit the frame. Lump has just hit the frame. Spring has its maximum distortion 30 cm a. Calculate the spring constant of the spring. b. For situation D, calculate the total kinetic energy K, the total gravitational potential energy Ugav], and the total elastic potential energy al Here, "total" means the sum of the quantities for the lump and the frame. Do the same for situation?. Same for situation ® Find the maximum distance, d, that the frame moves downward from its initial equilibrium position. c. d. e. f Show that mechanical energy is not conserved in the process O -0- -0. By hovw much did it change? Identify the irreversible physical phenomenon responsible for this change in mechanical energy

Explanation / Answer

a) use, F m*g

k*x = m*g

k = m*g/x

= 0.15*9.8/0.04

= 36.75 N/m

b) K1 = 0

Uel,1 = (1/2)*k*x^2

= (1/2)*36.75*0.04^2

= 0.0294 J

Ugrav,1 = m2*g*h

= 0.2*9.8*0.3

= 0.588 J

c) K2 = m2*g*h

= 0.2*9.8*0.3

= 0.588 J

Uel,2 = (1/2)*k*x^2

= (1/2)*36.75*0.04^2

= 0.0294 J

Ugrav,2 = 0

d) speed of putty just before hitting = sqrt(2*g*h)

= sqrt(2*9.8*0.3)

= 2.42 m/s

speed of both objects just after the collision,

v = 0.2*2.42/(0.2 + 0.15)

= 1.38 m/s

K3 = (1/2)*(0.2 + 0.15)*1.38^2

= 0.333 J

Uel3 = (1/2)*k*x^2

= (1/2)*36.75*0.04^2

= 0.0294 J

Ugrav,3 = 0

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