A block with a mass of 30.0 kilograms (kg) is held against a spring with a force constant of k = 1.25×105 N/m. The spring is initially compressed 17.5 centimeters (cm) from equilibrium. Then the block is released, setting it into motion. The block initially slides along a frictionless surface, but then it encounters a 1.50 meter long rough patch where the coefficient of kinetic friction is 0.134. It then slides down a frictionless ramp, which is tilted at an angle of 19.5 degrees and has a length of 2.50 meters. How fast is the block moving when it reaches the bottom of the ramp? Give your answer in meters per second (m/s). Hint: use WNC = ΔK + ΔU.

Q: A spring with spring constant 94 N/m and rest length 27.2 cm is used to pull a wooden block of…

A: GIven data The spring constant is given as k = 94 N/m. The rest length is given as l = 27.2 cm. The…

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Q: Problem 10: A block of mass m = 0.22 kg is set against a spring with a spring constant of k = 513…

A: mass (m) = 0.22 kg spring constant of first spring (k1) = 513 N/m Compression in first spring (x1) =…

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A: Given values m = 2 kg, X = 4.05 cm = 4.05 x 10-2 m Spring constant, k=590 N/m

Q: A spring of negligible mass stretches 3.46 cm from its relaxed length when a force of 5.40 N is…

A: The applied force F = 5.4 N. The particle’s mass is m = 0.8 kg. The amplitude A = 5 cm = 0.05 m. The…

Q: An 8.0-g bullet is suddenly shot into a 4.0-kg block that is at rest on a frictionless horizontal…

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Q: A 0.75 kg block is held at rest against a spring with a force constant k=730 N/m. The spring is…

A: Given data The mass of the block is: m = 0.75 kg The spring constant of the spring is: k = 730 N/m…

Q: A horizontal spring-mass system has low friction, spring stiffness 220 N/m, and mass 0.4 kg. The…

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A: The energy stored in spring will be transformed into the kinetic energy of the block when it reaches…

Q: A small block of mass M = 150 g is placed on top of a larger block of mass 3M which is placed on a…

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Q: what is its speed (in m/s) when it is a distance of 10 cm from the equilibrium position

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Q: A 2.85 kgkg block on a horizontal floor is attached to a horizontal spring that is initially…

A: Apply Newton’s second law at the vertical motion of the block, then n-mg=0n=mg=2.85 kg9.8 m/s2=27.93…

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A: The block's mass is, m=5 kg The spring constant is, k=130 N/m The magnitude of the force is, F=89 N…

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Q: A horizontal spring-mass system has low friction, spring stiffness 220 N/m, and mass 0.4 kg. The…

A: Given

Q: The block of mass m 1.60 kg slides along the horizontal rough plane as shown in the figure. The…

A: We solve this problem using conservation of energy

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A: Given: K=72 N/m F = 36 N work done by friction = 8.6 J initial height h = 2.0m

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Q: 18. A dart of mass 0.4 kg is pressed against the spring of a toy gun. The spring with a constant K =…

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A: Solution:

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Q: A horizontal spring with stiffness 0.8 N/m has a relaxed length of 16 cm (0.16 m). A mass of 22…

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Q: what is the velocity of the block vf in m/s?

A: Friction will perform negative work and decrease total mechanical energy of spring block system .…

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A: Solution: Given that package of mass m = 6.00 kg θ = 53.1 ° D = 4m  spring with force constant k =…

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A:

Q: A mass of 0.8 kg is attached to a horizontal spring whose stiffness is 10 N/m. Friction is…

A: Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for…

Spring Potential Energy

The potential energy is the energy possessed by a body by virtue of its position or the energy held by the object due to its position relative to other objects, its force like stresses, charge, and other factors affecting the particles of the body. The potential energy is the measure of the net work done by or on the body. The spring potential energy is the potential energy stored due to the deformation of an elastic spring and this elasticity is due to the stretched spring. The elasticity is the ability of a solid-state matter to come back to its equilibrium position or original position after any kind of external force is acted on it. It is also known as Elastic potential energy. The potential energy here is caused due to the displacement of the spring from its equilibrium position to another position and this potential energy is equal to the net work done due to the stretching of the spring. It also resembles the same mechanics of the oscillation of a simple pendulum, where due to the external force, the object moves from its equilibrium position to its maximum ends and the periodic motion continues till it comes back to the equilibrium position to rest.