Question

. (Serway 9th ed., 7-33) A 0.600-kg particle has a speed of 2.00 m/s at point A and a kinetic energy of 7.50 J at point B. What is (a) its kinetic energy at A, (b) its speed at B, and (c) the net work done on the particle by external forces as it moves from A to B? (Ans. (a) 1.20 J; (b) 5.00 m/s; (c) 6.30 J)

Answer 1

a) KA = 1.2 J

b) vB = 5.00 m/s

c) W = 6.30 J

Step-by-step explanation:

m = 0.600 kg

vA = 2.00 m/s

KB = 7.50 J

a) KA = ?

b) vB = ?

c) W = ?

We can apply the folowing equations

K = 0.5*m*v²

and

W = ΔK = KB- KA

then

a) KA = 0.5*m*vA² = 0.5*(0.600 kg)*(2.00 m/s)²

⇒ KA = 1.2 J

b) KB = 0.5*m*vB² ⇒ vB = √(2*KB / m)

⇒ vB = √(2*7.50 J / 0.600 kg)

⇒ vB = 5.00 m/s

c) W = ΔK = KB- KA = (7.50 J) - (1.2 J)

⇒ W = 6.30 J