Question

The answer to A) -0.883 J and the the answer to B) .285 N. I need help on how they got answer B. Thank you

4.) A small 60.0-g block is placed against a horizontal compressed spring with stiffness k- 360 N/m, at the bottom of a vertical, frictionless circular track (loop-the-loop) of radius R = 0.750 m, as shown below. The spring is initially compressed by 8.20 cm from its equilibrium length. smooth surface rough surface (a) [5 pts] What is the change in kinetic energy of the block as it slides from the bottom of the loop to the top? (b) [5 pts] What is the normal force between the block and the track at the top of the loop?

Answer 1

a] Change in kinetic energy of the block = potential energy gained by the block [from Energy conservation]

so, KE' - KE = - mgh = 0.06 x 9.8 x (2R) = - 0.882 J

b]

Kinetic energy at the bottom is:

a] Change in kinetic energy of the block = potential energy gained by the block [from Energy conservation] so, KE' - KE = -mgh = 0.06 times 9.8 times (2R) = -0.882 J b] Kinetic energy at the bottom is: KE = 1/2 kx^2 = 1/2 (360)(0.082)^2 = 1.2103J so, kinetic energy at the top is: KE' = 1.2103 - 0.882 = 0.3283 J the velocity is than, v = 3.308 m/s now, since the block is not falling off the top, the net force on it is zero N + mg = mv^2/R N = mv^2/R - mg = 0.287N this is the normal force on the block.

so, kinetic energy at the top is:

KE' = 1.2103 - 0.882 = 0.3283 J

the velocity is then, v = 3.308 m/s

now, since the block is not falling off the top, the net force on it is zero

$N+mg = \frac{mv^2}{R}$

$N = \frac{mv^2}{R} - mg = 0.287 \ N$

this is the normal force on the block.