wtons stretches a spring 4 meters. A mass of 40 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s of motion. m
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A: The potential energy stored converts to the kinetic energy. that is, 12kx2=12mv2kx2=mv2k=mv2x2
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A: mg=kx k=mgx
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A: Spring\'s Potential Energy(PE) is given by the equation :
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A: According to the conservation of energy, 12kx2=mghkx2=2mghk=2mghx2
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A: Given data: - The spring constant is K = 150 N/m. The mass is m = 0.25 kg. The distance travelled…
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A: We can estimate the spring constant using work energy theorem.
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A: The energy possessed by an object which is in motion , with velocity v is known as the kinetic…
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A: Given: Mass of block, m=4 kg Spring constant, k=20 N/m
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A: Givenk=50 N/mm=50 kgθ=30°x=0.3 mF=50 N
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A: M= 0.2kg Spring extends 4cm V= 0.68 m/s
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A: The potential energy stored in the spring is:
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Q: elastic potential energy stored in the spring
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A: Given:- Length of the gummy snake = 30 cm = 0.3 m Spring constant of the gummy snake = k = 15 N/m…
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Q: Q13: The block shown is released () from rest with the spring unstretched (). The pulley and the…
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A: mg=kx x=mg/k Total mechanical energy = Energy stored in spring + 12mv2 =12kx2+12mv2 = 12kmgk2+12mv2
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A: m=Mass of the gymnast=45 kgh=Height of the jump=0.6 mx=Displacement of the springk=Spring…
Q: A spring compresses 0.3 m and experiences a restoring force of 45 N. What is the spring constant?
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- A vibration sensor, used in testing a washing machine, consists of a cube of aluminum 1.50 cm on edge mounted on one end of a strip of spring steel (like a hacksaw blade) that lies in a vertical plane. The strips mass is small compared with that of the cube, but the strips length is large compared with the size of the cube. The other end of the strip is clamped to the frame of the washing machine that is not operating. A horizontal force of 1.43 N applied to the cube is required to hold it 2.75 cm away from its equilibrium position. If it is released, what is its frequency of vibration?A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. Give the initial conditions. x(0) = x'(0) = x(t) = m Find the equation of motion. m m/sA force of 720 newtons stretches a spring 3 meters. A mass of 60 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. Give the initial conditions. x(0)= m x′(0)= m/s Find the equation of motion. x(t) = m
- A mass of 18 kilograms stretches the spring 0.3 meters. Use this information to find the spring constant (use g = 9.81 m/s² as the acceleration of gravity). k = 588 The previous mass is detached from the spring and a mass of 14 kilograms is attached. This mass is displaced 0.7 meters below equilibrium and then launched with an initial velocity of 1 meters/second. Find the equation of motion. Note: When solving this problem, consider positions below equilibrium to be positive. x(t) =A force of 660 newtons stretches a spring 3 meters. A mass of 55 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 8 m/s. Find the equation of motion. x(t) = mA force of 720 newtons stretches a spring 4 meters. A mass of 45 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion. x(t) = -6- sin ( 2t)| x m
- If a spring is stretched 2.0 cm by a suspended object having a mass of0.55 kg, what is the force constant of the spring?A steel ball weighing 128 pounds is attached to a spring with spring constant of 6.25 Įb/ft. The ball is started in motion from the equilibrium position with an initial velocity of 6ft/sec in the downward direction. If the force due to air resistance is numerically equal to twice the instantaneous velocity, find the differential equation for the position of the ball at any time t.A 5.75-kg mass stretches a vertical spring 0.5 m. If the spring is stretched an additional 1.0 m and released, how long does it take to reach the (new) equilibrium position again? (What equation(s) can I use to solve this equation?)
- The diameter of a car's suspension spring decreases uniformly. The force exerted by the spring is given by F = axb. Consider a suspension spring that compresses x1 = 12.4 cm with a 1,000 N load and x2 = 34.0 cm with a 5,000 N load. (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.) (a) Find the constants a (in units of N/mb) and b. a= b= (b) Calculate the work (in J) needed to compress the spring 24.0 cm. (Assume the spring is not initially compressed.) JA mass weighting 24 lbs stretches a spring 2 inches. The mass is in a medium that exerts a viscous resistance of 28 lbs when the mass has a velocity of 2 ft/sec. Suppose the object is displaced an additional 6 inches and released. Find an equation for the object's displacement, u(t), in feet after t seconds. u(t) =A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 8 inches above the equilibrium position. Give the initial conditions. (Use g = 32 ft/s² for the acceleration due to gravity.) x(0) x'(0) x(t) = = = Find the equation of motion. ft ft ft/s