Question

1982m1. a 20 kg mass, released from rest, slides 6 meters down a frictionless plane inclined at an angle of 30° with the horizontal and strikes a spring of spring constant k = 200 newtons/meter as shown in the diagram above. assume that the spring is ideal, that the mass of the spring is negligible, and that mechanical energy is conserved. use g = 10 m/s², (sin30° = ½, cos 30° = 0.866)
a. determine the speed of the block just before it hits the spring.
b. determine the distance the spring has been compressed when the block comes to rest.
c. is the speed of the block a maximum at the instant the block strikes the spring? justify your answer.

Explanation:

Step1: Analyze forces and find acceleration

On a friction - less inclined plane, the net force along the plane is \(F_{net}=mg\sin\theta\). Using Newton's second law \(F = ma\), we have \(a=\frac{mg\sin\theta}{m}=g\sin\theta\). Given \(g = 10m/s^{2}\) and \(\theta = 30^{\circ}\), \(a=10\times\frac{1}{2}=5m/s^{2}\).

Step2: Use kinematic equation to find speed before hitting spring

The kinematic equation \(v^{2}=v_{0}^{2}+2ad\) is used. The block starts from rest (\(v_{0} = 0\)), \(a = 5m/s^{2}\), and \(d = 6m\). Substituting into the equation: \(v^{2}=0 + 2\times5\times6\). So \(v=\sqrt{60}\approx7.75m/s\).

Step3: Use conservation of mechanical energy for spring compression

The initial mechanical energy (gravitational potential energy at the start of compression) is converted into elastic potential energy of the spring. The initial height of the block with respect to the point of maximum compression is \(h=(6 + x)\sin30^{\circ}\), where \(x\) is the compression of the spring. The initial kinetic energy is \(\frac{1}{2}mv^{2}\) (from part a, \(v=\sqrt{60}m/s\)), and the elastic potential energy is \(\frac{1}{2}Kx^{2}\). Using conservation of energy \(mg(6 + x)\sin30^{\circ}+\frac{1}{2}mv^{2}=\frac{1}{2}Kx^{2}\). Substitute \(m = 20kg\), \(g = 10m/s^{2}\), \(K = 200N/m\), and \(v=\sqrt{60}m/s\):

$$20\times10\times(6 + x)\times\frac{1}{2}+\frac{1}{2}\times20\times60=\frac{1}{2}\times200\times x^{2}$$

$$100(6 + x)+600 = 100x^{2}$$

$$600+100x+600=100x^{2}$$

$$100x^{2}-100x - 1200=0$$

$$x^{2}-x - 12=0$$

Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 1\), \(b=-1\), \(c=-12\)), \(x=\frac{1\pm\sqrt{1+48}}{2}=\frac{1\pm7}{2}\). We take the positive root \(x = 4m\).

Step4: Determine if speed is maximum at strike

The speed of the block is not a maximum at the instant it strikes the spring. When the block strikes the spring, the spring starts to exert a force \(F=-Kx\) (opposing the motion of the block). The net force on the block is \(F_{net}=mg\sin\theta - Kx\). As long as \(mg\sin\theta>Kx\) (i.e., while the spring is still compressing and the net force is in the direction of motion), the block is still accelerating. The speed is maximum when \(mg\sin\theta=Kx\) (equilibrium point, not at the instant of strike).

Answer:

a. The speed of the block just before it hits the spring is \(\sqrt{60}\approx7.75m/s\).
b. The distance the spring has been compressed when the block comes to rest is \(x = 4m\).
c. No. Because when the block strikes the spring, the spring force \(Kx\) is less than \(mg\sin\theta\) (initially \(x = 0\), \(F_{spring}=0\)), and the net force \(mg\sin\theta\) is still in the direction of motion, so the block is still accelerating. The speed is maximum when \(mg\sin\theta=Kx\) (equilibrium point, not at the strike instant).