A 1.25 kg block slides on a frictionless, horizontal surface with an speed of 1.
ID: 1644831 • Letter: A
Question
A 1.25 kg block slides on a frictionless, horizontal surface with an speed of 1.37 m/sec. The block encounters an unstretched spring with a spring constant of 285 N/m. 1) What is the initial kinetic energy of the block before it hits the spring? KE_0 = 2) What is the potential energy of the mass and spring system when the spring is at its point of maximum compression? U_max = 3) How far is the spring compress before the block comes to rest? A = 4) If the mass and block were to oscillate, what would be the period of the oscillation? T = 5) How long is the block in contact with the spring before it comes to rest? t =Explanation / Answer
Given v = 1.37 m/s , m = 1.25 kg , k = 285 N/m
a) Kinetic energy = mv^2/2 = 1.25*1.37^2/2 = 1.173 joules
b) from the conservation of energy
Potential energy = 1.173 joules
c) from the conservation of energy
mv^2/2 = kx^2/2
1.173 = kx^2/2
x = (2*1.173/285)
x = 0.09 m
d) time period = 2(m/k)
T = 2(1.25/285)
T = 0.416 s
e) in this case,
T = 4t
t = T/4 = 0.416/4 = 0.104 s
t = 0.104 s
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