Problem_Set_8

Anthony Putrino Problem Set #8 – Problems 4, 12, 26, 36, 42, 46 4. A 2.8-kg block is attached to a spring with a force constant of 375 N/m. (a) Find the work done by the spring on the block as the block moves from A to B along paths 1 and 2. (b) How do your results depend on the mass of the block? Specifically, if you increase the mass, does the work done by the spring increase, decrease, or stay the same? (Assume the system is frictionless. (a) Solve using the sum of work done by the spring on path 1, and then path 2. Path 1: W = ½k[(x 1 2 -x 2 2 ) + (x 2 2 -x 3 2 )] = ½ (375N/m)[(0 2 - (.040m 2 )] + [(.040m 2 )-(.020m 2 )] = -.30J+.225J = -.075J Path 2: W = ½k[(x 1 2 -x 2 2 )+(x 2 2 -x 3 2 )] = ½ (375N/m)[0 2 -(-.020m 2 )] + [(-.020m 2 )-(.020m 2 ) = -.075J + 0J = -.075J (b) The work done is independent of mass of the block so the work done will remain the same whether the mass of the block is increased or decreased. 12. Experiments performed on the wing of a hawk moth ( Manduca sexta ) show that it deflects by a distance of x=4.8mm when a force of magnitude F=3.0mN F is applied at the tip. Treating the wing as an ideal spring, find (a) the force constant of the wing and (b) the energy stored in the wing when it is deflected. (c) What force must be applied to the tip of the wing to store twice the energy found in part (b)? (a) Solve using Hooke’s Law k = -(F/x) = -(.0030N/-.0048m) = .625N/m (b) Solve U using potential energy equation. U = ½ kx 2 = ½ (.625N/m)(.0048m) 2 = 7.2*10 -6 J (c) Solve using Hooke’s Law with doubling equation from potential energy equation (part b). 2U/k = x 2 converts to x = -sqrt(2U/k)

F = -kx = -k (-sqrt(2U/k) = sqrt (2Uk) **negatives cancelled* = sqrt (2(2*7.2*10 -6 J)(.625N/m) = .0042N 26. A 2.9-kg block slides with a speed of 2.1 m/s on a frictionless horizontal surface until it encounters a spring. (a) If the block compresses the spring 5.6 cm before coming to rest, what is the force constant of the spring? (b) What initial speed should the block have to compress the spring by 1.4 cm? (a) Calculate the force by using spring constant k. k = (mv 2 1 )/(x 2 max ) = (2.9kg)(2.1m/s)/.056m 2 = 4080N/m (b) Solve for v i . v i = sqrt((kx 2 max )/m) = sqrt((4030N/m)*(.014m 2 )/2.9kg) = .524m/s 36. Starting at rest at the edge of a swimming pool, a 72.0-kg athlete swims along the surface of the water and reaches a speed of 1.20 m/s by doing the work Wncl=+161J . Find the nonconservative work, W nc2 , done by the water on the athlete. Solve for the non consecutive work by equating the equation to the change in mechanical energy. W nc = W nc1 + W nc2 = W nc = K f - W nc1 = ½ mv f 2 - W nc1 = ½ (72kg)(1.20m/s) 2 - 161J = 109J 42. A 1250-kg car drives up a hill that is 16.2 m high. During the drive, two nonconservative forces do work on the car: (i) the force of friction, and (ii) the force generated by the car’s engine. The work done by friction is −3.11×10^5 J ; the work done by the engine is +6.44×10^5 J . Find the change in the car’s kinetic energy from the bottom of the hill to the top of the hill. Solve by setting the non consecutive work equal to the change in the mechanical energy to solve for the change in K.