A moving 4.30 kg block collides with a horizontal spring whose spring constant is 223 n/m. the block compresses the spring a maximum distance of 5.00 cm from its rest position. the coefficient of kinetic friction between the block and the horizontal surface is 0.340. what is the work done by the spring in bringing the block to rest? submit answer tries 0/12 how much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? submit answer tries 0/12 what is the speed of the block when it hits the spring?

1. Work done by spring = 0.279 Joules 2. Work lost due to friction = 0.716 Joules 3. Speed of block when first hit spring = 0.680 m/s 1. Using Hooke's law, the potential energy stored in the spring is E = 0.5kx^2 where E = potential energy k = spring constant x = distance the spring is deformed. Substitute the known values into the formula E = 0.5 223 N/m (0.05 m) ^2 E = 111.5 N/m 0.0025 m^2 E = 0.27875 Nm E = 0.27875 (kg m) / s^2 m E = 0.27875 (kg m^2) / s^2 E = 0.27875 J Rounding to 3 significant figures gives 0.279 Joules. 2. The amount of force needed due to kinetic friction is F = k * Fn where k = coefficient of friction Fn = Normal force The normal force is the mass of the object multiplied by the gravitational acceleration so, 4.3 kg * 9.8 m/s^2 = 42.14 (kg*m) / s^2 Now multiply by the coefficient of friction, getting 42.14 (kg*m) / s^2 * 0.340 = 14.3276 (kg*m) / s^2 = 14.3276 N So we have 14.3276 N over a distance of 5 cm (0.05m), so 14.3276 N * 0.05 m = 0.71638 Nm = 0.71638 J Rounding to 3 significant figures gives 0.716 Joules 3. The total work done on the block is the work used to compress the spring plus the work lost due to friction, so 0.279 J + 0.716 J = 0.995 J Now the energy of a moving object is expressed as the following equation. E = 0.5 M V^2 where E = Energy M = Mass V = Velocity. So setting energy equal to the amount used to stop the mass, we get 0.995 J = 0.5 M V^2 0.995 (kg*m^2) / s^2 = 0.5 M V^2 Substituting the known mass, getting 0.995 (kg*m^2) / s^2 = 0.5 4.3kg V^2 0.995 (kg*m^2) / s^2 = 2.15 kg V^2 And solve for V 0.995 (kg*m^2) / s^2 = 2.15 kg V^2 0.462790698 m^2/s^2 = V^2 0.680287217 m/s = V And finally, round to 3 significant figures, getting 0.680 m/s