A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the speed received in the above situation?Select one:a.1.73 db.2.00 dc.4.00 dd.1.41 d
Question
A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the speed received in the above situation?Select one:a.1.73 db.2.00 dc.4.00 dd.1.41 d
Solution
The kinetic energy of the mass is given by the equation KE = 1/2 * m * v^2, where m is the mass and v is the velocity. The potential energy stored in the spring is given by PE = 1/2 * k * x^2, where k is the spring constant and x is the compression of the spring.
In the first situation, the spring is compressed a distance d, so the potential energy in the spring is PE1 = 1/2 * k * d^2. This energy is converted into kinetic energy when the spring is released, so KE1 = PE1 = 1/2 * m * v^2.
In the second situation, we want the mass to attain double the speed, so the kinetic energy will be KE2 = 1/2 * m * (2v)^2 = 2 * KE1. To achieve this, we need to compress the spring a distance x such that PE2 = 1/2 * k * x^2 = KE2 = 2 * KE1 = 2 * 1/2 * k * d^2 = k * d^2.
Solving for x, we get x = sqrt(2) * d. Therefore, the compression of the spring that would result in the mass attaining double the speed received in the above situation is 1.41 times the original compression. So, the answer is d.1.41 d.
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