6. A 0.34 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.

Question

6. A 0.34 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B.
(a) What is its kinetic energy at A?

(b) What is its speed at point B?

(c) What is the total work done on the particle as it moves from A to B?


7. Find the height from which you would have to drop a ball so that it would have a speed of 8.6 m/s just before it hits the ground.

8. When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.19 cm.
(a) What is the force constant of the spring?

(b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it?

(c) How much work must an external agent do to stretch the same spring 7.70 cm from its unstretched position?

Explanation / Answer

6 mass of the object is m = 0.34 kg speed at A is vA = 5 m/s kinetic energy at B is KB = 7.5 J a) kinetic energy at A is KA = 0.5*mvA2 KA = 0.5*0.34 kg*(5 m/s)2 KA = 4.25 J b) speed of the object at B is vB = 2KB/m vB = 2*7.5 J/0.34 kg vB = 6.64 m/s c) According to work energy theorem, Work done = change in kinetic energy W = KB - KA W = 3.25 J 7 the final speed of the object just before it hits the ground is v = 8.6 m/s Let h be the initial height Then by conservation energy mgh = 0.5mv2 h = 0.5v2/g h = 0.5*(8.6 m/s)2/9.8 m/s2 h = 3.77 m 8 The mass of the object is m = 2.6 kg The weight of the object is w = 2.6 kg*9.8 m/s2 = 25.48 N The strectch in the spring is x = 3.19 cm a) According to Hooke's law F = kx k is the spring constant 25.48 N = k*3.19 cm k = 798.75 N/m b) If the mass is 1.3 kg 12.74 N =  798.75*x x = 1.59 cm c) The work done in stretching the spring is W = kx2 W = 798.75*(7.7 cm)2 W = 798.75*(7.7 cm)2 W = 47357.8875 J

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