![Horizontal Mass-Spring Systems Worksheet (Solutions)
Question 1:
A mass of 1.5kg, at rest, is attached to a spring, which is attached to a wall 75.0cm
away. An external force of 25.0N is required to compress the spring by 5.0cm.
Assuming the surface beneath the mass is frictionless, what is the spring constant k?
FNet = Fexternal + Fspring = 0
Fspring = -kx
0 = 25.0N + (-kx)= 25.0N – k (0.05m)
-25.0N/0.05m = - k
500N/m = k
Question 2:
A mass of 5.0kg, at rest on a frictionless surface, is attached to a spring, which is
attached to a wall 50.0cm away. An external force of 30.0N is used to stretch the
spring by 20.0cm. When the external force is removed: (A) What will the
acceleration of the mass be? (B) What will the angular frequency, the frequency and
the period of the objects motion be?
(A) FNet = Fspring = ma (B) = (k/m) k = -30.0N/-0.200m = 150N/m
Fspring = -Fexternal = ((150N/m)/(5.0kg)) = 30= 5.5 rad/s
-30.0N = ma f = /(2) = (5.5rad/s)/(2) = 2.8/s
-30.0N = (5.0kg)a T = 1/f = 1/(2.8/s) = 0.36s
-30.0N/5.0kg = a
-6.0m/s2 = a
Question 3:
A mass of 3.0kg is at rest on a frictionless surface, attached to a spring with a spring
constant of 26.0N/m. The spring is compressed by 30.0cm. When the mass is
released: (A) What will the amplitude of the motion be? (B) What will the phase
constant be?
(A) x(0)= Acos()
v(0) = -Asin()
[x(0)]2 = A2cos2()
[v(0)]2 = 2A2sin2()
(using trigonometric identity: 1 = cos2() +sin2())
2A2 = 2A2sin2() + 2A2cos2()
2A2 = 2[x(0)]2 + [v(0)]2
A2 = [x(0)]2+ [v(0)]2/2
A = {[x(0)]2+[v(0)]2/2} = {[-0.300]2+ [0.0m/s]2/2} = 0.300m](https://image.slidesharecdn.com/phys101lm2-learningobjectone-150126184556-conversion-gate01/85/Phys-101-LM2-learning-object-one-2-320.jpg)
![(B)(using the trigonometric identity tan() = sin()/cos())
2A2tan()= 2A2sin()/2A2 cos()
tan()= [v(0)]/2[x(0)]
= tan-1[(0.00m/s)/2(-0.300)] = 0](https://image.slidesharecdn.com/phys101lm2-learningobjectone-150126184556-conversion-gate01/85/Phys-101-LM2-learning-object-one-3-320.jpg)
Phys 101 LM2 learning object one
- 1. Horizontal Mass-Spring Systems Worksheet Question 1: A mass of 1.5kg, at rest, is attached to a spring, which is attached to a wall 75.0cm away. An external force of 25.0N is required to compress the spring by 5.0cm. Assuming the surface beneath the mass is frictionless, what is the spring constant k? Question 2: A mass of 5.0kg, at rest on a frictionless surface, is attached to a spring, which is attached to a wall 50.0cm away. An external force of 30.0N is used to stretch the spring by 20.0cm. When the external force is removed: (A) What will the acceleration of the mass be? (B) What will the angular frequency, the frequency and the period of the objects motion be? Question 3: A mass of 3.0kg is at rest on a frictionless surface, attached to a spring with a spring constant of 26.0N/m. The spring is compressed by 30.0cm. When the mass is released: (A) What will the amplitude of the motion be? (B) What will the phase constant be?
- 2. Horizontal Mass-Spring Systems Worksheet (Solutions) Question 1: A mass of 1.5kg, at rest, is attached to a spring, which is attached to a wall 75.0cm away. An external force of 25.0N is required to compress the spring by 5.0cm. Assuming the surface beneath the mass is frictionless, what is the spring constant k? FNet = Fexternal + Fspring = 0 Fspring = -kx 0 = 25.0N + (-kx)= 25.0N – k (0.05m) -25.0N/0.05m = - k 500N/m = k Question 2: A mass of 5.0kg, at rest on a frictionless surface, is attached to a spring, which is attached to a wall 50.0cm away. An external force of 30.0N is used to stretch the spring by 20.0cm. When the external force is removed: (A) What will the acceleration of the mass be? (B) What will the angular frequency, the frequency and the period of the objects motion be? (A) FNet = Fspring = ma (B) = (k/m) k = -30.0N/-0.200m = 150N/m Fspring = -Fexternal = ((150N/m)/(5.0kg)) = 30= 5.5 rad/s -30.0N = ma f = /(2) = (5.5rad/s)/(2) = 2.8/s -30.0N = (5.0kg)a T = 1/f = 1/(2.8/s) = 0.36s -30.0N/5.0kg = a -6.0m/s2 = a Question 3: A mass of 3.0kg is at rest on a frictionless surface, attached to a spring with a spring constant of 26.0N/m. The spring is compressed by 30.0cm. When the mass is released: (A) What will the amplitude of the motion be? (B) What will the phase constant be? (A) x(0)= Acos() v(0) = -Asin() [x(0)]2 = A2cos2() [v(0)]2 = 2A2sin2() (using trigonometric identity: 1 = cos2() +sin2()) 2A2 = 2A2sin2() + 2A2cos2() 2A2 = 2[x(0)]2 + [v(0)]2 A2 = [x(0)]2+ [v(0)]2/2 A = {[x(0)]2+[v(0)]2/2} = {[-0.300]2+ [0.0m/s]2/2} = 0.300m
- 3. (B)(using the trigonometric identity tan() = sin()/cos()) 2A2tan()= 2A2sin()/2A2 cos() tan()= [v(0)]/2[x(0)] = tan-1[(0.00m/s)/2(-0.300)] = 0