A 0.018 kg block sits on a frictionless surface, inclined at an angle of 22 degrees. The mass is resting on a spring with a spring constant of 12.6 N/m. The block is pushed to compress the spring 0.1 meters and then is let go. How fast is the block moving when it loses contact with the spring?

Question

killemall · Answer

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We don't care about forces normal to the surface.  The equation for the total force is this:

$${F} = -k\textrm{x} + \textrm{mg} \sin(22)$$

This means that the potential energy is:

$$PE = 0.5 * k\textrm{x}^2 - \textrm{mgx} \sin(22)$$

The spring loses contact with the block as soon as the block is moving faster than the spring.  They have the same velocity from x=-0.1 to x=0, but as soon as the spring and block goes past x=0, they have different accelerations and therefore, different velocities. So we just need to calculate their velocities at x=0.  At x=0, all energy is kinetic, so we calculate the potential energy at x=-0.1 and equate it to 1/2mv^2

$$PE=0.5 * 12.7 * (-0.1)^2 -(0.018)(-9.8)(-0.1) \sin(22) = 0.0569 J$$
$$v=\sqrt{2*PE/m}$$
velocity = 2.51 m/s