A 4.1-kg bowling ball is held 10 mm above a mattress and then released from rest

Question

A 4.1-kg bowling ball is held 10 mm above a mattress and then released from rest and allowed to fall, sinking into the mattress. Model the mattress as a single spring with spring constant k = 330 N/m .

Calculate the ball's kinetic energy at the following compressions: 0, 0.050 m, 0.10 m, 0.15 m, 0.20 m

Calculate the gravitational potential energy of the system (ball, spring, and Earth) at the following compressions: 0, 0.050 m, 0.10 m,0.15 m, 0.20 m. Measure the gravitational potential energy relative to the height of the uncompressed bed.

Calculate the elastic potential energy at the following compressions: 0, 0.050 m, 0.10 m, 0.15 m, 0.20 m.

What is the magnitude of the maximum compression of the mattress? Ignore the inertia of the spring.

Explanation / Answer

here,
m = 4.1 kg
H = 10 mm = 0.001 m
k = 330 N/m

Case1 : x = 0.050 m
Case2 : x = 0.10 m
Case3 : x = 0.15 m
Case4 : x = 0.20 m

A)
Solving for KE :

from Conservation of Energy Ke = potential Energy gained by spring
KE = 0.5 * k * x^2

Case1 : x = 0.050 m
KE = 0.5 * k * x^2
KE = 0.5 * 330 * 0.050^2
KE = 0.4125 J

Case2 : x = 0.10 m
KE = 0.5 * k * x^2
KE = 0.5 * 330 * 0.10^2
KE = 1.65 J

Case3 : x = 0.15 m
KE = 0.5 * k * x^2
KE = 0.5 * 330 * 0.15^2
KE = 3.7125 J

Case4 : x = 0.20 m
KE = 0.5 * k * x^2
KE = 0.5 * 330 * 0.20^2
KE = 6.6 J

B)
Solving for Gravitational potential Energy :
Height of ball when uncompressed bed = 0.001 m

GPE = mass * acceleration due to gravity * Height from uncompressed bed

GPE = 4.1 * 9.8 * 0.001
GPE = 0.040 J

This GPE will remain same for all case when bed is uncompressed.

C)
Elastic potential Energy :

Case1 : x = 0.050 m
U = 0.5 * k * x^2
U = 0.5 * 330 * 0.050^2
U = 0.4125 J

Case2 : x = 0.10 m
U = 0.5 * k * x^2
U = 0.5 * 330 * 0.10^2
U = 1.65 J

Case3 : x = 0.15 m
U = 0.5 * k * x^2
U = 0.5 * 330 * 0.15^2
U = 3.7125 J

Case4 : x = 0.20 m
U = 0.5 * k * x^2
U = 0.5 * 330 * 0.20^2
U = 6.6 J

D)
Magnitude of maximum compression can be found by using :

0.5 * k * x^2 = m * g *(h + x)

0.5*330*x^2 = 4.1*9.8(0.001+x)
165*x^2 = 40.18x + 0.004018

by solving this qudratic equation we get :

Xmax = 0.2445 m

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