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Well, it's not energy, its power $P$. $~~~~~~~~~~~~P = \int F \cdot dv$ And since power is the derivative of energy $P = \dot E$, your world makes sense again ;). Regarding your problem I agree with Yanping Cai, the kinetic energy of the car $E_{kin}$ must be converted into potential energy of the spring $E_{spring}$. $~~~~~~~~~~~~\frac{1}{2} m_{max} ... 1 This answer may not look like a typical answer, but I am attempting to instill a key concept, so please bear with me. For this type of problem, where you are investigating a possible solution, UNITS are EXTREMELY important. What are the units of momentum? What are the units of force? Note that if units do not match across an equal sign, the answer is ... 1 I'm sure you can consider the force and integrate it over displacement to calculate the total work has been done. But why don't you step back and consider the conservation of energy? In short, $$\dfrac{1}{2}mv^{2} =\dfrac{1}{2}kx^{2}$$$k$is the minimum spring constant it requires. 0 If the mass was simply attached at the current position and let free, your energy method is correct. If the mass was attached by hand and slowly allowed to stretch the spring until it remains fixed at a stable equilibrium point, the force method would have been correct. However, for your given situation I believe the question is asking you for the first ... 0 The two approaches do apply to different situations. Let's first look at the "energy method". Here you describe what happens to the different energy components as the mass starts to fall from$x=h$. The initial energy is indeed$mgh$. Now, after that the mass has fallen a distance$x$, the energy is composed of three components: the residual potential energy ... 0 Your force considerations are wrong. Note that there is only force on the block (i.e.$kx$); but there are actually two forces which are extending the spring. One is being applied by the block hanging(which you have considered). But you have not considered the force on the spring due to it's attachment to the ceiling. There will be an additional equal ... -1 The answer is no. The net force at the maximum elongation points has the same magnitude. This is because the rest point of the spring is modified by the gravitational weight of the mass. The mass oscillates around this new rest point, and at the points of maximum amplitude the net force is the same. One can tell that the gravitational force can be ... 2 In accordance with Hooke's law the force is linear with distance. Incorporating gravity only means that the equillibirum position of the spring has changed, the "zero" around which it oscillates. The gravitational pull is already compensated by the spring. Thus the magnitude of the force is euqal at$-A$and$+A$. Edit: When the gravitational pull on the ... 5 The equilibrium position in this case is not where the spring is not stretched, it is actually stretched by a$\Delta x$amount with$F_{spring}(0) = k\Delta x$. So the spring force on point A is a little smaller than in point -A, since$ F_{spring}(A) = -k(A-\Delta x)$and$ F_{spring}(-A) = k(A+\Delta x)$so it compensates the "extra" force. You have ... 0 If you take away mass in Newtonian mechanics, then any force leads to an infinite response. The spring force then promises an infinite negative-feedback response to any deviation from equilibrium, and depending on how you take the limits involved, you either get infinite sinusoidal motion with a period of 0, or perfect rigidity with no motion and a period of ... -1 If a force f (Net force) acts on the mass(Spring Balance), it will accelerate with a = f/ m. Spring balance will only show the reading if there is net force causing deflection of spring (F= K*x). Here f = Net Force on Spring causing acceleration F= Force that causes deflection of spring(Which always acts in pair - equal in magnitude and opposite in ... 0 In terms of representing the answer: F would be for the reading because that is the basic notation for total force, and as there is no need to differentiate between the two forces Left or Right due to them being equal, then only F should be used. The actual answer as you probably will know is F = 0. When L is not equal to R, the reading should be along the ... 0 Control moment gyros , like those used on the space station can store tremendous amounts of energy and are used to create torque to change spacecraft attitude (rotation). The energy is stored by the wheel's high angular velocities, and the torque is created by tilting the wheel (pushing against the wheel's angular momentum). 1 There are four ways that I can think of to store mechanical (elastic) energy. Axial Stretch${\rm d}W = \frac{F^2}{2 E A} {\rm d}x$Torsion${\rm d}W = \frac{T^2}{2 G J} {\rm d} x$Bending${\rm d}W = \frac{M^2}{2 E I} {\rm d} x$Shearing${\rm d}W = \frac{c F^2}{2 G A} {\rm d} x\$ See https://en.wikipedia.org/wiki/Castigliano%27s_method and last page in ...