Block, spring, and mass problem

The stone is at maximum displacement when the block is at its equilibrium position (no external net force acting on it). Thus, the net force on the stone at this point is equal to the force of gravity acting on it, which is 0.028 kg * 9.8 m/s^2 = 0.2744 N.
  • #1
kiwikahuna
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0

Homework Statement


A 0.1 kg block is suspended from a spring. When a small stone of mass 28 g is placed on the block, the spring stretches an additional 4.4 cm. With the stone on the block, the spring oscillates with an amplitude of 14 cm.
(a) What is the frequency of the motion?
(b) How long does the block take to travel from its lowest point to its highest point?
(c) What is the net force on the stone when it is at a point of maximum upward displacement?



The Attempt at a Solution


I was able to solve a & b, but am having trouble with c.

First, I found the spring constant by using the info that the 28 g stretched the spring 4.4 cm.

F = k Δx or k = F / Δx = mg / Δx = 0.028 * 9.8 / 0.044 = 6.236 N/m

Then I found the angular freq of the motion:

ω = ( k / m )1/2 = ( 6.236 / 0.128 )1/2 = 6.98 rad/s

and then the freq is

f = ω / 2π = 6.98 / 2 π = 1.11 Hz

The period of the motion is

T = 1/f = 1 / 1.11 Hz = 0.900 sec so the time to get just from bottom to top is half this or

0.450 sec

Finally, I thought that the net force on the block at the top is

F = kA = 6.236 N/m * 0.14 m = 0.873 Newtons but this answer isn't right. What did I do wrong though?


Please help! thank you!
 
Last edited:
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  • #2
I think your spring constant is wrong. With only the 0.1 kg block on the spring the spring stretches by an amount
[tex]
\Delta x = \frac{-m_{1} g}{k}
[/tex]
where [tex]m_{1}[/tex] is the mass of the block. When the stone is placed on the block the spring stretches an additional 4.4 cm. This would give

[tex]
0.044(meters) + \Delta x = \frac{-(m_{1}+m_{2}) g}{k}
[/tex]

Where [tex]m_{2}[/tex] is the mass of the stone. I would use these two equations to solve for [tex]k[/tex].

When both masses are on the spring, this would lead to the following equation of motion:

[tex]
(m_{1}+m_{2}) \ddot{x} = -k x - (m_{1}+m_{2}) g
[/tex]

Where [tex]x[/tex] is the displacement from equilibrium when there is zero mass on the spring.
 
Last edited:
  • #3
Your spring constant is just fine. The question asks for the net force on the stone, not on the block. It may help to find the acceleration of the block (and hence the stone) at the point.
 

1. What is a block, spring, and mass problem?

A block, spring, and mass problem is a type of physics problem that involves a block attached to a spring and a mass. The block is typically placed on a horizontal surface with a spring attached to one side and a mass attached to the other. The goal of the problem is to determine the block's motion and any forces acting on it.

2. What is the equation for the force of a spring?

The equation for the force of a spring is F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position. This equation is known as Hooke's Law and describes the relationship between the force applied to a spring and its resulting displacement.

3. How do you calculate the period of oscillation for a block, spring, and mass system?

The period of oscillation for a block, spring, and mass system can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. This equation assumes that there is no friction or other external forces acting on the system.

4. How does the mass affect the motion of a block, spring, and mass system?

The mass of the block affects the motion of the system by changing the period of oscillation. A larger mass will result in a longer period, meaning the block will take longer to complete one full oscillation. The mass also affects the amplitude of the oscillation, with a larger mass resulting in a smaller amplitude and vice versa.

5. How is energy conserved in a block, spring, and mass problem?

In a block, spring, and mass problem, energy is conserved because the system is isolated and there are no external forces acting on it. This means that the total mechanical energy (kinetic + potential) remains constant throughout the motion of the block. The energy is constantly being converted between kinetic and potential energy as the block oscillates back and forth.

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