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A 1.35 kg block at rest on a tabletop is attached to a horizontal spring having constant 19.8 n/m. the spring is initially unstretched. a constant 23.8 n horizontal force is applied to the object causing the spring to stretch. the acceleration of gravity is 9.8 m/s 2 . 23.8 n 1.35 kg 19.8 n/m find the speed of the block after it has moved 0.236 m from equilibrium if the surface between block and tabletop is frictionless. answer in units of m/s.

User Erel
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Final answer:

The speed of the 1.35 kg block after it has moved 0.236 m from equilibrium on a frictionless surface is calculated using the conservation of energy principle. The work done by the applied force is equated to the sum of the kinetic energy and the elastic potential energy in the spring.

Step-by-step explanation:

To find the speed of the 1.35 kg block after it has moved 0.236 m from equilibrium on a frictionless surface with a spring constant of 19.8 N/m, we can use the conservation of energy principle. The work done by the external force transforms into the kinetic energy of the block and the potential energy stored in the spring.

First, we calculate the work done by the force (W) using the equation W = F × d, where F is the force applied (23.8 N) and d is the distance moved (0.236 m). Then, we determine the potential energy (PE) stored in the spring as PE = (1/2)kx^2, where k is the spring constant (19.8 N/m) and x is the distance the spring is stretched from its equilibrium position (0.236 m).

Since there's no friction to dissipate energy, the work done by the force goes into the kinetic energy (KE) of the block and potential energy of the spring. So, we have W = KE + PE. The kinetic energy of the block can be found using the equation KE = (1/2)mv^2, where m is the mass of the block and v is its velocity. Solving for v gives us the speed of the block.

User Mottek
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horizontal force is applied to the object causing the spring to stretch. the acceleration of gravity is 9.8 m/s
User Adrianos
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