A block of mass (2 kg) starts at rest on a plane inclined (35 degrees) above the horizontal. The kinetic coefficient of friction between the block and the plane is (0.15). After being released, the block travels a distance of (2.3 m) down the plane before reaching the bottom of the plane. At the bottom of the inclined plane is a horizontal spring with spring constant of (7 N/m). How far is the spring compressed when the block collides with it?

Q: A vertically mounted spring (k = 850 N/m) is compressed by 24.0 cm relative to its unstrained…

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A: mass of ball = 2.03 kg k = 1108 Nm x = 0.201 m coefficient of kinetic friction = 0.249d = 1.44 m θ =…

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Q: A block of mass m = 0.32 kg is set against a spring with a spring constant of k1 = 681 N/m which has…

A: Using the conservation of energy, 12k1d12-μkmgx=12k2d22

Q: As seen in Fig. 1, a 5.00 kg mass starts from rest at the top of a frictionless incline at heighty0=…

A: Work done by gravitational force will be = mgh Work = 5 x 9.8 x 3 = 147 J  Using Work energy…

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Q: A 1.2 kg block is held at rest against a spring with a force constant k = 750 N/m. Initially, the…

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Q: A block with mass 0.50kg is forced against a horizontal spring of negligible mass, compressing the…

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Q: 3.0 kg k= 400 N/m 3.4 m

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A: m = 0.220 kg K = 120 N/m x1 = 11.2 cm = 0.112 m

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Q: A mass m = 12 kg rests on a frictionless table and accelerated by a spring with spring constant k =…

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

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A: It is given that, m=1.9 kgk=500 N/mx=7 cm=0.07 m

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A: We know,

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A: according to the conservation of energy, Kf-Ki+Uf-Ui=0                               1

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Q: A box is pressed against a horizontal spring, compressing the spring from its relaxed length. The…

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Q: A box of mass 15.0-kg is released from position 1. The path is frictionless except for the portion…

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Q: A block of mass m = 2.5 kg is attached to a spring with spring constant k = 890 N/m. It is initially…

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Q: A 0.35 kg mass is held against a compressed spring with a spring constant of 89 N/m. The mass is…

A: Using the work-energy theorem, Ki+Ui+Wnc=Kf+Uf0+12kx2-μmgx=12mv2+0v=kx2m-2μgx

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Q: A block of mass m = 1.5 kg is attached to a spring with spring constant k = 510 N/m. It is initially…

A: 1. Draw the schematic arrangement of the system.

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.