Two trains start from towns 224 mi apart and travel towards each other on parallel tracks. They pass each other 1.6 hr later. If
1 answer:
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Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.
W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.
Answer:
(a) vmax = 0.34m/s
(b) v = 0.13m/s
(c) v = 0.31m/s
(d) x = 0.039m
Explanation:
Given information about the spring-mass system:
m: mass of the object = 0.30kg
k: spring constant = 22.6 N/m
A: amplitude of the motion = 4.0cm = 0.04m
(a) The maximum speed of the object is given by the following formula:
(1)
w: angular frequency of the motion.
The angular frequency is calculated with the following relation:
(2)
You replace the expression (2) into the equation (1) and replace the values of the parameters:
The maximum speed of the object is 0.34 m/s
(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:
The speed is 0.13m/s
(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:
You solve the previous equation for t:
With this value of t, you can calculate the speed of the object with the following formula:
The speed of the object for x = 1.5cm is v = 0.31 m/s
(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:
The position will be:
The position of the object on which its speed is one-half its maximum velocity is 0.039