Hooks law: F=-x*K
F=Force
x=distance = 0.5m
K=constant
F=mass*9.81 = 30*9.81=aaaa [N]
aaa = -0.5 * K => K = something
The spring constant can be calculated using Hooke's Law which states F = kx, where F is the force applied, k is the spring constant, and x is the displacement. In this case, the force is equal to the weight of the object, so F = m * g, where m is the mass and g is the acceleration due to gravity. Plugging in the values, k = (m * g) / x = (30kg * 9.8m/s^2) / 0.5m = 588 N/m.
Answer: Spring constant = 500 N/m
The spring constant is calculated by dividing the weight of the object (29 N) by the distance it stretches the spring (11 cm). First, convert 11 cm to meters by dividing by 100 (0.11 m), then divide the weight by the stretch distance to get the spring constant: 29 N / 0.11 m = 263.6 N/m.
The spring stretches more at the top when suspended because the gravitational force acting on the spring is greater at the top due to the weight of the spring itself. This leads to larger elongation at the top compared to the bottom of the spring.
The spring constant is calculated using Hooke's Law: F = kx. For a 2-N weight, k = F / x = 2 N / 0.04 m = 50 N/m. Using this constant, the spring stretches for the 6-N weight is x = F / k = 6 N / 50 N/m = 0.12 m.
A spring scale measures weight by measuring the force necessary to stretch or compress a spring in the device. The more the spring stretches or compresses when an object is hung from it, the greater the force or weight of the object. This force is then translated into a weight reading on the scale.
The spring scale got its name because it uses a spring to measure force or weight. The spring within the scale stretches or compresses based on the applied force, allowing for accurate measurement of the weight of an object.
The oscillation of a spring is the motion that the spring makes when disturbed. Imagine holding the end of a spring and hanging a weight to the other end. If you do not disturb the weight, it will stay in a static position. However, when you pull down on the weight and let go, the spring "oscillates" up and down. The spring could also be compressed and released, creating the same effect. The up and down motion, which has a specific velocity and period relating to the spring constant k, is oscillation.
The spring constant is calculated using Hooke's Law: F = kx. For a 2-N weight, k = F / x = 2 N / 0.04 m = 50 N/m. Using this constant, the spring stretches for the 6-N weight is x = F / k = 6 N / 50 N/m = 0.12 m.
If a 2.5N extends the spring for 0.1m, so a 7.5N will extend 3 times that, a 0.3m.
Scales used to measure weight include: * Spring scale - measures the increased length of a spring as it stretches * Balance scale - uses a horizontal lever to compare unknown weights to that of known weights
No, the length of the string does not affect the reading of a spring scale. The scale measures the force applied to it, which is not influenced by the length of the string.
see me as earth
A spring scale measures weight. It is dependent on the gravity, and so one's weight in space, or on the moon would be different based on the spring scale. A balance measures mass, and since the standards in a balance are the same no matter what the gravity is, then the mass remains constant. However, for all practical purposes, the gravity on earth is constant so the weight and mass of the objects would be the same.
A spring scale measures weight. It is dependent on the gravity, and so one's weight in space, or on the moon would be different based on the spring scale. A balance measures mass, and since the standards in a balance are the same no matter what the gravity is, then the mass remains constant. However, for all practical purposes, the gravity on earth is constant so the weight and mass of the objects would be the same.
The net force acting in the stretch direction on the spring is proportional to its deformation (e.g., its stretch). In math talk that's F = kdX where k = 50 N/m and F = mg = 40*9.81 is the weight of the m mass. NOTE: 40 kg is not...not...a weight, it's a measure of mass. To get the weight, the force of gravity, we must multiply by g = 9.81 m/sec^2 which is a typical average acceleration due to gravity on the Earth's surface. So the stretch dX = F/k = mg/k = 40*9.81/50 = 7.848 meters. ANS.
If actually weighing the plates is impractical, you could try hanging the plates from a spring, and testing to find the spring's k value, and recording the displacement of the object while hanging from the spring, and use that to calculate the force on the plate, which equals mg. if the density is known, you could immerse the plates in something to find their volume and then calculate their weight from that. or, you could try and pull them with a force meter, taking two data points so that you can solve for the both the friction coefficient and weight.
Yes, a spring will stretch when a weight is added to it due to the force of gravity acting on the weight, causing the spring to deform. The amount of stretch will depend on the weight added and the stiffness of the spring.
The spring constant (k) determines the stiffness of the spring and should ideally remain constant for a specific spring. If you are obtaining different k values for the same spring using different masses, there may be errors in your measurements or variations in the spring itself such as deformation or wear over time. It is important to ensure that your measuring equipment is accurate and that the spring is not damaged.