A block with mass 0.80 kg moves on a horizontal frictionless surface. The block

Question

A block with mass 0.80 kg moves on a horizontal frictionless surface. The block is attached to one end of a horizontal spring and the other end of the spring is attached to the wall. When the block is at x = +0.30 m, its speed is 6.0 m/s and the magnitude of its acceleration is 12.0 m/s2. What is the maximum kinetic energy of the block during its motion and at what value of x does this occur? Ans. Kmax = 15.8j x= 0 What is the maximum value for the acceleration of the block during its motion and at what value of x does this occur? Ans. amax= 39.8 m/s2 x= plusminus 0.994 m

Explanation / Answer

1.At X = +0.3m we can write

F = kx = ma ( k = Spring constant)

k*0.3 = 0.8*12 on solving we get k = 32 N/m

maximum kintetic eneregy is achieved when the sping unstretched which is at x = 0.

The total energy at x =3 gets converted in to kinetic energy at x = 0.

Energy at x = 3 is 0.5*m*v^2+0.5*k*0.3^2

which is 15.84J i.e max kinetic energy.

2.The maximum acceleration is achieved at maximum extension i.e. when ll kinetic is converted in to potential energy. Total energy is 15.84.

At maximum extention 15.84 = 0.5*k*x^2

on solving we get x = 0.994m

apply k*x = M*a solving this we get a(max) = 39.8 m/s^2

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