Solving Spring Problems Involving Mass and Frictionless Floor

In summary, the problem involves a block with mass 4.5 kg attached to a spring with relaxed length 3 m and spring constant 16 N/m. The block is pulled 3 m to one side and is always constrained to move on the frictionless floor. The spring extends by 3 m and the potential energy stored is 12.3 J. The maximum speed the block attains is 2.34 m/s. In part (d), the block is given an initial speed equal to 1.8 times the speed found in part (c), and it travels a distance of 4.287 m before stopping. However, in part (e), the magnitude of the acceleration is 0.638 m/s
  • #1
Naeem
194
0
A block of mass m = 4.5 kg rests on a frictionless floor. It is attached to a spring with a relaxed length L = 3 m. The spring has spring constant k = 16 N/m and is relaxed when hanging in the vertical position. The block is pulled d = 3 m to one side. In this problem, the block is always constrained to move on the floor (i.e. it never leaves the floor).


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a) By what amount is the spring extended?
DL = m *


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b) What is the potential energy stored in the spring?
Uspring = J *
12.3 OK


--------------------------------------------------------------------------------
c) The block is released but is constrained to move horizontally on the frictionless floor. What is the maximum speed it attains?
|v|max= m/s *
2.34 OK


--------------------------------------------------------------------------------
Let's change the problem a bit. When the spring is vertical (hence, unstretched), the block is given an initial speed equal to 1.8 times the speed found in part (c).
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d) How far from the initial point does the block go along the floor before stopping?
Dmax = m *
4.287 OK


--------------------------------------------------------------------------------
e) What is the magnitude of the acceleration of the block at this point (when the spring is stretched farthest)?
|a| = m/s2
0.638 NO

HELP: What is the force exerted by the spring on the block when the spring is fully stretched?

I am not able to figure out part e. I'm stuck

here is what I did F = ma = k * delta L

Plugged in m , k and delta L

found a to be 8.25 m/s2, which is wrong, anybody tell me what is wrong?
 
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  • #2
The force a spring exerts on an object is F = -kx.
 
  • #3
If the spring is vertical in its relaxed mode, and the block is move laterally, then the spring forms the hypotenuse of an isosceles right triangle with two sides of 3 m.

Here I am assuming the floor is horizontal with the normal parallel to the axis of the spring.
 
  • #4
Can anybody help me with part e) here please.

Thanks,
 

1. How do I calculate the force required to compress a spring with a given mass on a frictionless floor?

The force required to compress a spring can be calculated using Hooke's Law, which states that the force applied to a spring is directly proportional to the distance it is stretched or compressed. The formula is F = -kx, where F is the force, k is the spring constant, and x is the distance the spring is compressed. With a frictionless floor, the force required will be equal to the weight of the mass, so the formula becomes F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2).

2. How does the mass of an object affect the compression of a spring on a frictionless floor?

The mass of an object does not affect the compression of a spring on a frictionless floor. According to Hooke's Law, the force required to compress a spring is directly proportional to the distance it is compressed, not the mass of the object compressing it. This means that a heavier object will compress the spring the same amount as a lighter object, as long as they are both on a frictionless surface.

3. What is the relationship between the spring constant and the stiffness of a spring on a frictionless floor?

The spring constant is a measure of the stiffness of a spring, so the higher the spring constant, the stiffer the spring will be. This means that a spring with a higher spring constant will require more force to compress it a certain distance, while a spring with a lower spring constant will be easier to compress.

4. Does friction have any effect on the compression of a spring on a frictionless floor?

No, friction does not have any effect on the compression of a spring on a frictionless floor. Friction is a force that resists motion, but since there is no friction on a frictionless floor, it will not affect the compression of a spring. However, if the floor is not completely frictionless, the force required to compress the spring will be slightly higher due to the friction between the floor and the object.

5. Can a spring on a frictionless floor be compressed indefinitely?

No, a spring on a frictionless floor cannot be compressed indefinitely. The spring will reach a point where it can no longer be compressed any further, known as the spring's elastic limit. This is because the spring's material will eventually reach its maximum capacity to store elastic potential energy and any additional force applied will cause permanent deformation, also known as plastic deformation.

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