A spring has spring constant k, and is initially compressed by a distance A. Upo

Question

A spring has spring constant k, and is initially compressed by a distance A. Upon release, the spring launches a block of mass m along a horizontal frictionless surface. The incline has diagonal length L and angle theta, and is FRICTIONAL. The coefficient of kinetic friction between the block and the incline is mu_k,. Finally, the block gets to the top of the incline and flies off in projectile motion. Draw force diagrams of the block when it is attached to the spring, on the incline and in the air (3 diagrams total). In each picture, indicate the direction of acceleration. Compute the horizontal distance x_f from the upper edge of the incline that the block will land. Assume that m, k, A, mu_k and L are all known. Now imagine the sequence of events is reversed, so that a cannon launches a projectile from the position found in part b), and the box eventually smashes into the spring and becomes attached. What is the angular frequency of the resulting oscillation?

Explanation / Answer

When attached to spring, Only force is F = kA towards right.

When on the incline,

Normal force perpendicular to incline. Tangential force is friction which is (mu ) * N

And vertically downward force is mg.

When in projectile motion, the only force is mg vertically down

(b)From conservation of energy

We have (0.5 k A^2 ) - (mu) * N * L = mgLsin (theta) + (0.5 m v^2)

So v = sqrt [ {(0.5 k A^2 ) - (mu) * mg cos(theta) * L - mgLsin (theta)} / 0.5 *m]

Now From equations of motion,

- L sin(theta) = v sin ( theta)t - 0.5gt^2; Substituting v = we can calculate t

and x = vcos(theta) * t; Substitute t to calculate x

(c)

The box should be launched with same final velocity to retrace the path of projectile

We know from conservation of energy that

(0.5 k A^2 ) - (mu) * N * L = 0.5 m v (final ) ^2

This is the velocity with which it hits the ground. So this should be the velocity of launch to retace the path.

Again the block looses energy due to friction

So Before getting attached the final energy would be

(0.5 k A^2 ) - (mu) * N * L*2   = 0.5* k* (new amplitude) ^2

The angular frequency is independent of Amplitude. It is dependent on spring since T = 2*pi * sqrt ( m/k)

So f = (1/2*pi) sqrt ( k/m)

so omega = sqrt (k/m) which is dependent on mass and spring constant, not the amplitude.

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