You use a slingshot to launch a 40g stone upward at an angle (not straight upwar

Question

You use a slingshot to launch a 40g stone upward at an angle (not straight upward) The spring has a spring constant of 100 N/m and an unstretched length of 10 cm. To launch the stone you stretch the elastic to 30 cm. At the highest point in its flight, the stone moves at5m/s. Ignore friction and air resistance, and assume the initial height of 1.7 m. Full points will be awarded for the correct answer to the following question: What is the height of the highest point in the flight? If the full credit cannot be awarded, then partial credit will be awarded for the correct answers to the following questions: i) What is the initial stretch of the elastic? ii) what is the initial potential energy of the stretched elastic? iii) what is the initial kinetic energy of the stone before elastic is released? iv) What is the gravitational energy of stone before the elastic is released? v) what is the total energy of the system just before the elastic is released? vi) what is the total energy of the system once the stone is at the highest point? vii) why is the kinetic energy of the stone at the highest point of trajectory is NOT zero? viii) what is the kinetic energy of the stone at the highest point? ix) what is the potential energy of the dart when it is the highest point? x) What is the height of the dart at the highest point?

Explanation / Answer

A) By energy conservation,

0.5kx^2 = 0.5mv^2 + mgh

0.5*100*0.3^2 = 0.5*0.040*5^2 + 0.040*9.8*h

h = (0.5*100*0.3^2-0.5*0.040*5^2) /(0.040*9.8)

= 10.2 m

Total height =10.2+1.7 = 11.9m answer.

i) 0.30 m

ii) SPE =0.5*100*0.3*0.3=4.5J

iii)KE=0

iv)GPE=0.040*9.8*1.7 = 0.6664 J

v) TE =4.5+0.6664= 5.1664 J

vi) TE =5.1664 J

vii) because it has non zero velocity.

viii) KE = 0.5*0.040*5^2 = 0.5J

ix) PE = 5.1664-0.5 = 4.6664J

x) h = 4.6664/(0.040*9.8) = 11.9 m

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