Exatas
A mass of 3 kg falls freely a distance of 5 m before contacting a spring attached to the ground. If the spring stiffness is 400 N/m, what is the maximum spring compression?
Given: Data on mass and spring Find: Maximum spring compression Solution The given data is M 3 kg h 5m k 400 N m
Apply the First Law of Thermodynamics: for the system consisting of the mass and the spring (t spring has gravitional potential energy and the spring elastic potential energy) Total mechanical energy at initial state E1 Mgh
Total mechanical energy at instant of maximum compression xE2
M g ( x)
1 2 kx 2
Note: The datum for zero potential is the top of the uncompressed spring But E1 E2 M g ( x) 1 2 kx 2 0
so
Mgh
Solving for x
x
2
2Mg x k
2Mgh k
x
Mg k
Mg k
2
2Mgh k
x
3 kg
9.81
m s
2
m 400 N m s
2
3 kg
9.81
m 400 N
2
2
3 kg 9.81
m s
2
5m
m 400 N
x
0.934 m
Note that ignoring the loss of potential of the mass due to spring compression x gives 2Mgh k
x
x
0.858 m
Note that the deflection if the mass is dropped from immediately above the spring is 2Mg k
x
x
0.147 m
Problem 4.4
Problem 4.5
Problem 4.7
Problem 4.8
Given: Data on cooling of a can of soda in a refrigerator Find: How long it takes to warm up in a room Solution The First Law of Thermodynamics for the can (either warming or cooling) is dT dt dT dt k Mc T is the
Mc
k T
Tamb
or
A T
Tamb
where
A
where M is the can mass, c temperature, and Tamb is the ambient temperature dT T Tamb T ( t) Tamb
Separating variables
A dt
Integrating
Tinit
Tamb e
At
where Tinit is the initial temperature. The available data from the coolling can now be used to ob a value for constant A Given data for cooling Tinit ( 25 273) K Tinit 298 K
Tamb
(5
273) K
Tamb
278 K
T
( 10
273) K
T
283 K
when
t
10 hr
Hence