Problem Set 8

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SUNY Empire State College *

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1200

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Mechanical Engineering

Date

Feb 20, 2024

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docx

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Uploaded by DeaconMosquitoMaster1223

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. Given: M= 2.8kg K= 375 N/m (a) To solve the work done by path 1 and 2 we would follow as 0.5 (375N/m) [(0^2- (.040m^2)] + [(.040m^2) - (.020m^2)] First bracket = 0^2−0.040m^2= −0.040m^2 Second bracket = 0.040m^2−0.020m^2=0 .020m^2 Now calculate the terms separately: 0.5 ⋅ (375N/m) x (−0.040m2) + 0.020m2 1. First term: 0.5 x (375 N/m) x (−0.040 m2) =−7.5 Joules 2. Second term: 0.020 m2=0.020 Joules Now, let's add these two terms together: −7.5 Joules+0.020 Joules=−7.48 Joules = -.075J The negative sign indicates that the work is done in the spring (the spring is compressed) as it moves from the initial to the final position. (b) The work done by the spring does not depend on the mass of the block. As you can see, in both cases (path 1 and path 2), the work done by the spring is determined solely by the spring's force constant and the displacement of the block . Increasing the mass of the block will not change the work done by the spring.

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)F = 0.0030N x = 0.0048m Hooke’s Law = k = -(F/x) -(.0030N/-.0048m) = .625N/m (b)To solve U or the energy stored in the wing when deflected we would use the potential energy equation as U = 0.5 kx^2 0.5 x (0.625N/m) (0.0048m) ^2 = 0.00001152m2 ⋅ (0.625N/m) = 0.0000072Joules = 7.2 x 10^-6 Joules (c)To solve the force applied to store twice the energy, we would use Hooke’s law and doubling the energy equation. It would be shown as = Sqrt (2 (2 x 7.2 x 10^-6 Joules) (0.625N/m) = 0.042 N, So, the force required to store twice the energy is 0.042N 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)To solve (a) we would calculate the force by potential energy in the spring as: m is the mass of the block (2.9 kg). k = force constant of the spring e = extension (in meters) = 0.056m^2 v is initial speed of block = 2.1m/s (2.9kg) (2.1m/s) ^2 / 0.056m^2 = 4080N/m

(b) We would solve initial speed that the block compress by 1.4cm which would follow as: We want to compress the spring by 0.014 m (1.4 cm), so =0.014m^2 Sqrt ((4030N/m) x (0.014m^2) / 2.9kg) = 0.52m/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. m is the mass of the athlete (72.0 kg). v is the final velocity (1.20 m/s). 0.5 (72kg) (1.20m/s) ^2 - 161J = 109J So, the nonconservative work done by the water on the athlete is 109 Joules, and the positive sign indicates that work is done on the athlete by the water. 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. We would set the nonconservative work equal to the change in mechanical energy as: (6.44 x 10^5 J) + (-3.11x 10^5 J) – (1250 kg) (9.81 m/s^2) (16.2 m) 1.34 x 10^5 J = 134 kJ is the change in the car’s energy from bottom of hill to top Friction slows down the car, but the engine makes it go faster, so the car gains more energy, making it faster when going uphill. 46. An object moves along the x axis, subject to the potential energy. The object has a mass of 1.1 kg and starts at rest at point A. (a) What is the object’s speed at point B? (b) At point C? (c) At point D? (d) What are the turning points for this object? (a) The kinetic energy for B is 2 J and the kinetic energy + potential energy = 10 J We would calculate as Sqrt (2 x 10J – 2J) / 1.1kg = 3.8m/s is the speed for point B (b) Sqrt (2 x 10 J – 6J) / 1.1kg = 2.7 m/s (c) Sqrt (2 x 10J – 5J) / 1.1kg = 3.0 m/s

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