A mass of 1. 0 kg is pushed horizontally against a massless spring with a spring

Question

A mass of 1. 0 kg is pushed horizontally against a massless spring with a spring constant of 25 N/m until the spring compresses 0. 20 w, If the mass is then released what is the kinetic Energy of the mass when it is no longest in contact with the spring. (Ignore Friction) J. 15. A tumtable votates Freely, without a motor, and without Friction or air resistance, with a moule sitting at it's center. The angular velocity of the tumtable with the masse is: w, It's angular momentum is: L, and it's moment of Inertia is I. The mouse now walks to the edge of the tumtable. As the masse does so, the combination of the mouse and the tumtable, a) w, L, and I all increase b) w increases, but L and I remain the same; c) w and I increase but L remains the same; d) w decrease, L remains the same, and I increases; e) w and L decrease, but I increases;

Explanation / Answer

14)

spring constant ( k ) = 25 N/m

distance compressed ( x ) = 0.20 m

mass ( m ) = 1.0 kg

for a spring mass system, Kinetic energy = spring potential energy

P.E = K.E

0.5*k*x2 = 0.5*m*v2

v = 1 m/s2

15)

From newton's 2nd law, momentum is always conserved in a system unless acted upon by an external force

so since the mouse and table are considered as a system together, the angular momentum( L ) remains the same.

But moment of inertia is generally represented as mass*distance2, where distance is the distance from the axis of rotation.

Since the mouse move away from the axis, the moment of inertia of the mouse increases, so the total moment of inertia of mouse+table( I ) increases.

Momentum is given as I*w

since it remains same, angular velocity( w ) should decrease since I increases.

Option D

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