A spring of spring constant k = 8.25 N/m is displaced from equilibrium by a distance of 0.150 m.?
The formula for stored energy is E = Kx²/2 E = 8.25*0.15²/2 = 0.0928 joule
A horizontal spring attached to a wall has a force constant of k = 820 N/m.?
This is an energy balance problem since there is no friction. Change in energy of the system = 0 Change in energy of the system = 1/2*k*(xf^2 - xi^2) + 1/2*m*(vf^2 - vi^2) a) When the spring is stretch/compressed by 5cm all the energy is stored as potential in the spring. Total energy = 1/2*k*x^2 = 1/2* 820 N/m * (.05 cm) ^2 = 1.025 J b) When it passes through the equilibrium point, all the spring energy is transferred into kinetic energy and the velocity is max 1.025J = 1/2 * m * v^2 v = sqrt(2* 1.025 J / m ) = 1.26 m/s c) At xi/2 = 2.50 cm you have part of the energy still in spring potential and part in kinetic energy Remember, the change in energy = 0 = 1/2*k*(xf^2 - xi^2) + 1/2*m*(vf^2 - vi^2) 1/2*k*(xi^2 - xf^2) = 1/2 * m * (vf^2 - vi^2) vi = 0 m/s 1/2*820 N/m * ( (.05 m)^2 - (.025 m)^2 ) = 1/2* 1.30 kg *vf^2 vf = sqrt(820 N / m / 1.30 kg * (.0025 m^2 - .000625 m^2) ) = 1.09 m/s
A 0.20 kg object, attached to a spring with a spring constant k = 10 N/m, is moving on a horizontal frictionl?
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A spring of spring constant k = 8.25 N/m is displaced from equilibrium by of 0.150 m.?
W=1/2(8.25)(0.150)=0.61875N
Basic physics spring question: is my answer correct?
The gravitational force (ie weight) reqd to displace A) spring 0.230 m is: F = kx = 6.50(0.230) = 1.495 N the mass of a weight = 1.495 N = 1.495/g = 1.495/9.81 = 0.152 kg ANS A) √ The gravitational force of a 1 kg mass = mg = (1)(9.81) = 9.81 kg F = kx 9.81 = 12.5x x = 0.785 m ANS B) √ to 2 digit significance both of your answers are correct :>) however, U might want to notice that the data in this problem was given to 3 significant digits
A spring has a stiffness constant k = 150 N/m. A mass m = 4.0 is released from rest when the spring is stretch?
1) Total energy of the system is always constant. So, initially, when the spring is maximally stretched, the body has no kinetic energy and only potential energy, so total energy at any point of time = potential energy at start = 0.5kx^2 = 0.5 x 150 x 0.64 = 48 J 2) Total energy at x = x0/2 is Kinetic energy + Potential energy. Potential energy at that instant is 0.5k(x0/2)^2 = 0.5 x 150 x 0.16 = 48/4 J = 12 J So kinetic energy = Total energy - potential energy = 48 - 12 = 36 J 3) Maximum KE = 0 potential energy + Maximum kinetic energy = total energy = 48 J 4) You need to find v in 0.5mv^2 = 48 i.e. v = 4.9 m/s 5) Maximum acceleration occurs at extreme position and it is given by (x0)w^2 where w^2 = k/m = 150/4 = 37.5 So acceleration = 0.8 x 37.5 m/s^2 = 30 m/s^2
Equivalent Spring Constant (Series)When putting two springs in their equilibrium positions in series attached at the end to a block and then displacing it from that equilibrium, each of the springs will experience corresponding displacements x1 and x2 for a total displacement of x1 + x2. We will be looking for an equation for the force on the block that looks like:The force that each spring experiences will have to be same, otherwise the springs would buckle. Moreover, this force will be the same as Fb. This means thatWe can solve for :Now we just plug this back into Fb:Now, we equate Fb = F1:Canceling x1 we get:Which can also be written:Source: Series and parallel springsAnd so you have:[math]\frac{1} {k_{eq}}=\frac{1}{k}+\frac{1}{k}+\frac{1}{k}=\frac{3}{k}[/math]
Quick physics question (Mass on a Spring)?
I was given two physics questions and I figured out the answer except I'm unsure as to whether or not they should be positive or negative. 1. A spring has a spring constant k=8.75 N/m. If the spring is displaced 0.150 m from its equilibrium position, what is the force that the spring exerts? I used F=kx and got 1.31 N but I'm not sure if I was supposed to use F= -kx instead. 2. A spring of constant k=11.75 N/m is hung vertically. A 0.500 kg mass is suspended from the spring. What is the displacement of the end of the spring due to the weight of the .500 kg mass? I used x=(mg)/k and got x=0.417 m. Here, I have the same problem where I don't know if the equation should contain a variable with a negative value instead, giving me a negative answer. Can someone explain if the values of my answers should be positive or negative?
Mass of object m =75 g = 0.075 kgWeight of the object = mg = 0.075*10 N (Taking g=10 m/s²)=0.75 NDifference in length = 7–4 = 3 cm =0.03 mSpring constant k = 0.75/0.03 N/m = 75/3 N/m =25 N/mTotal displacement from 4.0 cm to 10 cm = 6 cm = 0.06 mHence total work done against spring force =½kx² =½*25*0.06*0.06 J 0.045 J
Any physicist knows that if an object applies a force to a spring, then the spring applies an equal and opposite force to the object. Hooke’s law gives the force a spring exerts on an object attached to it with the following equation:F = –kxwhere the minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. (k is called the spring constant, which measures how stiff and strong the spring is. x is the distance the spring is stretched or compressed away from its equilibrium or rest position.)The force exerted by a spring is called a restoring force; it always acts to restore the spring toward equilibrium. In Hooke’s law, the negative sign on the spring’s force means that the force exerted by the spring opposes the spring’s displacement. Hooke’s law saysF = –kxLooking only at the magnitudes and therefore omitting the negative sign, you get