a block of wood of mass m1= 1 kg slides to the right on a frictionless horizonta

Question

a block of wood of mass m1= 1 kg slides to the right on a frictionless horizontal table with a speed of 30 m/s. it collides with a second piece of wood of mass m2 = 2kg which is stationary at position X. The collission is completely inelastic. after the collision, the wood slides across a rough horizontal surface (friction component is .3) from B to C for distance 10m. After position C the surface becomes smooth and the wood collides with a spring constnat of k=30.00 N/m/

A) how much is the spring compressed by the wood?

B) at what locations along the horizontal plane is mass M2 stationary?

Please so work so I can follow. Thanks!

Explanation / Answer

Hooke's law says that F = -kx, where x = distance the spring has been stretched or compressed from equilibrium F = is the restoring force exerted by the spring k = the spring constant (with units of force/length) So if the spring was compressed to 20.0 cm instead of 4.00 cm, it will have the maximum F_2 will be 5 times as great as before, i.e. F_2 = 5F. Also, the average force will be 5 times as great (the force decreases linearly over the distance as it returns to the equilibrium position). But the kinetic energy of the wood would be the same as the potential energy when the spring is at it's maximum compression. Pe_2 = F_2*d_2 = F_2*x_2. Since this F_2 = 5F and x_2 = 5x, this gives PE_2 = 25*F*x. So the answer is K_2 = 25 * K. Or you could do it more directly by knowing that the potential energy of a spring is given by the formula PE = (1/2)kx^2 Since x_2 = 5x, PE_2 = (1/2)kx_2^2 PE_2 = (1/2)k(5x)^2 PE_2 = 25*(1/2)kx^2 PE_2 = 25*PE Since the potential energy is 25 times as great, and since it is all converted to kinetic energy, K_2 = 25 K.

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