A spring with a spring constant k = 1600 N/m is at rest on the bottom of an incl

Question

A spring with a spring constant k = 1600 N/m is at rest on the bottom of an inclined plane. A 7.0-kg block slides down the plane and makes contact with the spring at point A as shown After contact, the spring is compressed to point B, 0.20 m from point A where the speed of the block is zero m/s. What was of the speed of the block just before contact with the spring? When a 0.20-kg block is suspended from a vertically hanging spring, it stretches the spring from its original length of 0.040 m to 0.045 m. The same block is attached to the same spring placed on a horizontal, frictionless surface as shown. The block is then pulled so that the spring stretches to a total length of 0.15 m. The block is released at time t = 0 s and undergoes simple harmonic motion. What is the frequency of the motion?

Explanation / Answer

5.)
Elastic potential energy of spring when block is stationary = 1/2kx2
= 0.5 x 1,600 x 0.20 x 0.20 = 32 J
When the block just makes contact with the spring, elastic potential energy of spring = 0
Kinetic energy of block = 1/2mv2 = 0.50 x 7.0 x v2 = 3.5 v2 J
During the movement of the spring, gravitational potential energy lost by the block = mgh
= 7.0 x 9.8 x 0.20 sin 55o = 11.2 J
Using conservation of energy, 3.5 v2 = 32 + 11.2 = 43.2
v2 = 43.2/3.5 = 12.34
v = 12.34 =) 3.5 m/s or 2.4 m/s

6.)
k = F/d = mg/d N/m
w = sqrt(k/m) = sqrt(g/d) = 42.272 rad/s (using g = 9.8 m/s^2)
f = w/(2pi) = 7.0641 Hz
Max displ. A = 0.15 m
Max accel. = w^2A = gA/d = 294 m/s^2

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