You use a spring-loaded pistol to launch a 0.04-kg dart up at an angle. The spri

Question

You use a spring-loaded pistol to launch a 0.04-kg dart up at an angle. The spring has a spring constant of 500 N/m and an un-stretched length of 10 cm. Before launch, the spring is compressed to a length of 4 cm. At the highest point of the dart’s arc, it is moving at 5 m/s. Set the initial height of the dart to 0 m. Ignore air resistance. How much potential energy is in the spring once it is compressed?What is the kinetic energy of the dart at the top if its arc? What is the potential energy of the dart at the top of its arc?What is the height of the dart at the top of its arc?

Explanation / Answer

Given that :

mass of the dart, m = 0.04 kg

spring constant, k = 500 N/m

un-stretched length of the spring, x1 = 10 cm = 0.1 m

stretched length of the spring when compressed, x2 = 4 cm = 0.04 m

initial velocity of dart, v0 = 5 m/s

initial height of the dart, h0 = 0 m

(a) The potential energy required when spring compressed which will be given as :

P.Espring = (1/2) k x2                                                                            { eq.1 }

inserting the values in above eq.

P.Espring = (0.5) (500 N/m) [(0.1 m) - (0.04 m)]

P.Espring = (250 N/m) (0.06 m)

P.Espring = 15 J

(b) The kinetic energy of the dart at the top if its ar which is given as :

K.E = (1/2) m v02                                                                { eq.2 }

inserting the values in eq.2,

K.E = (0.5) (0.04 kg) (5 m/s)2

K.E = 0.5 J

(c) The potential energy of the dart at the top of its arc which is given as :

using conservation of energy, we have

P.Etop = K.Etop + P.Espring                                                                { eq.3 }

inserting the values in eq.3,

P.Etop = (0.5 J) + (15 J)

P.Etop = 15.5 J

(d) Height of the dart at the top of its arc which will be given as :

P.Etop = m g htop                                                              { eq.4 }

where, g = acceleration due to gravity = 9.8 m/s2

inserting the values in eq.4,

(15.5 J) = (0.04 kg) (9.8 m/s2) htop     

htop = (15.5 J) / (0.04 kg) (9.8 m/s2)

htop = (15.5 J) / (0.392 N)

htop = 39.5 m

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