A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits a spring with stiffness k at the end of the track causing the spring to compress. What is the maximum compression x of the spring? By conservation of energy, K1 + U1 = K2 + U2 Some of the terms in the equation above are zero; thus it can be simplified to: V½m 2 = 2 Isolating the compression x, X = ( 21 1/2 Suppose the mass of the toy car moving at 2.5 m/s is 0.0505 kg and the spring constant isk = 605 N/m, the spring will experience a maximum compression of 0.0 84 m.

Q: A 0.15 kg ball is dropped. How fast will it be moving after it has fallen 1.5 meters? (Note – there…

A: Mass of the ball is m = 0.15 kg  hi = 1.5 m

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A: SOlution:

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A: Given quantities: mass of the car = m speed of the car = v1 spring constant = k maximum compression…

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A: According to conservation of energy Total energy of isolated system remains constant

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A:

Q: A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle…

A:

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A: The initial kinetic energy of a toy car can be given as, Initially, the spring is not compressed or…

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A:

Q: QUESTION 5 k |m₁ = ? Vo m₂ Xmax A first mass is at rest on a frictionless horizontal surface and…

A: the compression is maximum when the masses comes to rest

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A:

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A:

Q: A spring gun (k= 28.0 N/m) is used to shoot a 41.3-g ball horizontally. Initially the spring is…

A: Given: The spring constant is 28 N/m. The mass of the ball is 41.3 g. The initial compression in the…

Q: A box of mass m is pressed against (but is not attached to) an ideal spring of force constant k and…

A: Given that Mass of the box is = m Spring constant is = k Initial spring compressing distance is = x…

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Q: A 30.2 N net force is applied for 15.64 s to a 10.2 kg box initially at rest.  What is the speed of…

A: Using Newton's second law, the acceleration of the box can be obtained as…

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A: By conservation of energy,  K1+U1=K2+U2 Here, K1 and K2 are the initial and final kinetic energies…

Q: A 6.00 ✕ 105 kg subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large…

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Q: What is the maximum compression x of the spring?

A: As per the conservation law of energy. Here, E, K, U, m, h, v, k, and x represent the car’s total…

Q: spring with a constant of 50.0 N/cm. The launcher is prepared by compressing the spring by 5.00 cm.…

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Q: In the figure, a block of mass m = 15 kg is released from rest on a frictionless incline of angle 0…

A: (a) Draw the free-body diagram of the block.

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A: Given data: Mass of toy car is, m=0.0505 kg. Velocity is, v1=2.5 m/s. Spring constant is, k=605 N/m.…

Q: A 7.5-kg object passes through the origin at time t = 0 such the its x component of velocity is…

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Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A: Given that: Mass of the toy car = m Initial speed of the car= v1  Spring constant= k Let the…

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A: By conservation of energy, K1+U1=K2+U2 Initially , the spring is in the relaxed position so that the…

Q: A force acting on a 2.00 kg object is given by F(x) = (2.00 N/m) x + (1.00 N/m³)x³. An object starts…

A: Given value--- mass = 2 kg . F = 2x + x3 . We have to find--- speed of object at x = 2 m .

Q: A box of mass m = 0.12 kg is set against a spring with a spring constant of k1 = 595 N/m which has…

A: Write the given values with the suitable variables. m=0.12 kgk1=595 N/md1=0.1 mk2=158 N/m

Q: A block of mass 15.0 kg slides from rest down a frictionless 33.0° incline and is stopped by a…

A: mass, m = 15 kg  Angle, A = 33°  k= 4 x 104 N/m distance, d= 3 m

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A: given: spring constant = K compressed distance = x velocity just as it moves free to spring = v…

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Q: A spring of a spring gun, having constant k- 628 N/m, ia compressed horizontally a distancex-0.05 m…

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Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

A: The problem can be solved using the conservation of energy.

Q: A force acting on a 2.00 kg object is given by F(x) = (2.00 N/m)x+ (1.00 N/m³)x³. An object starts…

A: The mass of the object is m=2.00 kg The force acting on the mass is Fx=2x+1x3

Q: A toy car of mass m is pushed along a frictionless track so that it is moving at a speed v1. It hits…

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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.