A block having a mass of 0.72 kg is given an initial velocity V_A = 1.2 m/s to t

Question

A block having a mass of 0.72 kg is given an initial velocity V_A = 1.2 m/s to the right and collides with a spring whose mass is negligible and whose force constant is k = 49 N/m as shown in the figure. Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision. Suppose a constant force of kinetic friction acts between the block and the surface, with mu_k = 0.49. If the speed of the block at the moment it collides with the spring is V_A = 1.2 m/s, what is the maximum compression X_C in the spring? Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision. The various parts of the figure help us imagine what the block will do in this situation. All motion takes place in a horizontal plane, so we do not need to consider changes in gravitational potential energy. We identify the system to be the block and the spring. The block spring system is isolated with no nonconservative forces acting. Before the collision, when the block is at point A, it has kinetic energy and the spring is uncompressed, so the elastic potential energy stored in the system is zero. Therefore, the total mechanical energy of the system before the collision is just 1/2 mv_A^2. After the collision, when the block is at point C, the spring is fully compressed; now the block is at rest and so at rest and so has zero kinetic energy. The elastic potential energy stored in the system, however, has its maximum value 1/2 kx^2 = 1/2 kx^2_max, where the origin of coordinates x = 0 is chosen to be the equilibrium position of the spring and x_max is the maximum compression of the spring, which in this case happens to be x_C. The total mechanical energy of the system is conserved because no nonconservative force act on objects within the isolated system. Write a conservation of mechanical energy equation: K_C + U_pC = K_A + U_pA 0 + 1/2 Kx_max^2 = 1/2 mV_A^2 + 0 Solve for x_max and evaluate: x_max = squareroot m/K V_A = squareroot 0.72 kg/49 N/m (1.2 m/s) = m Suppose a constant force of kinetic friction acts between the block and the surface, with mu_x = 0.49. If the speed of the block at the moment it collides with the spring is V_A = 1.2 m/s, what is the maximum compression x_C in the spring? Because of the friction force, we expect the compression of the spring to be smaller than in part (A) because some of the block's kinetic energy is transformed to interval energy in the block and the surface. We identify the system as the block, the surface, and the spring. This system is isolated but now involves a nonconservative force. In this case, the mechanical energy E_mech = K + U_ of the system is not conserved because a friction force acts on the block. From the particle in equilibrium model in the vertical direction, we see that n = mg. Evaluate the magnitude of the friction force: f_R = mu_x n = mu_x mg Write the change in the mechanical energy of the system due to friction as the block is displaced from x = 0 to x_c: delta E_mech = -f_R x_C Substitute the initial and final energies: delta E_mech = E_r - E_i = (0 + 1/2 kx_c^2) - (1/2 mV_A^2 + 0) = -f_k x_C = 1/2 kx_c^2 - 1/2 mv_A^2 = -mu_k mgx_c Substitute numerical values: 1/2 (49) x_c^2 - 1/2 (0.72)(1.2)^2 = -(0.49)(0.72)(9.80)x_c 24.5 x_c^2 + 3.5 x_c - 0.52 = 0 The physically meaningful root of this quadratic equation is x_c = m The negative root does not apply to this situation because the block must be to the right of the origin (positive value of x) when it comes to rest. Notice that the value for x_c is less than the distance obtained in the frictionless case of part (A) as we expected. The spring is now mounted vertically on the table, and the mass is dropped downwards, hitting the spring and compressing it. Just before the 'collision', the block has a measured velocity of 2.40 m/s downwards. What will be the maximum compression of the spring? (Friction is negligible.) m

Explanation / Answer

Master it question :-
k = 49 N/m
m = 0.72 kg

Let the compression in the spring be x
Using Enrgy conservation,
Kinetic Energy of mass + Gravitational Potential Energy of mass = Spring Potential Energy of mass
1/2 * m * v^2 + m*g*x = 1/2*k*x^2
1/2 * 0.72 * 2.4^2 + 0.72 * 9.8 * x = 1/2 * 49 * x^2
x = 0.468 m

Maximum compression of the spring, x = 0.468 m

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