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kupik [55]
3 years ago
8

A mass, M, is at rest on a frictionless surface, connected to an ideal horizontal spring that is unstretched. A person extends t

he spring 30 cm from equilibrium and holds it at this location by applying a 10 N force. The spring is brought back to equilibrium and the mass connected to it is now doubled to 2M. If the spring is extended back 30 cm from equilibrium, what is the necessary force applied by the person to hold the mass stationary there
Physics
1 answer:
nignag [31]3 years ago
7 0

Answer:

The necessary force applied by the person to hold the mass stationary there is 10 N

Explanation:

We are told that this person extends the spring 30 cm from equilibrium and holds it at this location by applying a 10 N force.

Thus, based on Hooke's law formula which is F = k Δx, we can say that the mass attached to the spring does not change the spring constant. Thus, the

same resistive spring force will still be in place and in turn, the same stretching force of 10N would still be required.

Thus;

The necessary force applied by the person to hold the mass stationary there is 10 N

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So,

Moment of inertia (I_y_y )  of the system with respect to the axis yy.

⇒ I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g

⇒ I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2

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