A lengthy question, I know. please help me answer all parts of this question. th
ID: 2040271 • Letter: A
Question
A lengthy question, I know. please help me answer all parts of this question. thank you very much.
2. A 2 kg object is attached to a spring and is oscillating on a horizontal surface. When the object has a speed of 10 m/s, the spring is stretched 2 m. The spring constant is 10 N/m. Neglect friction Find the maximum speed of the object. a. b. What is the maximum stretch in the spring? c. What is the objeer' sped when the spring is stretched I m? d. What is the stretch in the spring when the object's speed is 5 m/s? A slingshot has a spring constant of 50 N/m. A 0.005 kg marble is placed in the slingshot and the slingshot is stretched 0.2 m. If energy is conserved, then what maximum speed will the slingshot give the marble? e. f. A spring, with k 20 N/m, is stretched 0.5 m horizontally. A 0.5 kg mass is attached to the spring and released. Ignoring friction, what is the maximum speed of the mass.Explanation / Answer
2)
given
m = 2 kg
v = 10 m/s, at x = 2m
k = 10 N/m
a) Apply conservation of energy
(1/2)*m*v_max^2 = (1/2)*m*v^2 + (1/2)*k*x^2
v_max^2 = v^2 + (k/m)*x^2
v_max = sqrt(v^2 + (k/m)*x^2)
= sqrt(10^2 + (10/2)*2^2)
= 11 m/s
b) (1/2)*k*x_max^2 = (1/2)*m*v_max^2
x_max^2 = (m/k)*v_max^2
x_max = sqrt(m/k)*v_max
= sqrt(2/10)*11
= 4.92 m
c) Apply conservation of energy
(1/2)*m*v^2 + (1/2)*k*x^2 = (1/2)*m*v_max^2
(1/2)*2*v^2 + (1/2)*10*1^2 = (1/2)*2*11^2
==> v = 10.8 m/s
d) Apply conservation of energy
(1/2)*m*v^2 + (1/2)*k*x^2 = (1/2)*m*v_max^2
(1/2)*2*5^2 + (1/2)*10*x^2 = (1/2)*2*11^2
==> v = 4.38 m
e)
Apply conservation of energy
(1/2)*m*v_max^2 = (1/2)*k*x_max^2
v_max = sqrt(k/m)*x_max
= sqrt(50/0.005)*0.2
= 20 m/s
f)
Apply conservation of energy
(1/2)*m*v_max^2 = (1/2)*k*x_max^2
v_max = sqrt(k/m)*x_max
= sqrt(20/0.5)*0.5
= 3.16 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.