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A slingshot consists of a light leather cup containing a stone. The cup is pulled back against two parallel rubber bands. It takes a force of 15 N to stretch either one of these bands 1.0 cm. (a) What is the potential energy stored in the two bands together when a 58-g stone is placed in the cup and pulled back 0.10 m from the equilibrium position?
A slingshot consists of a light leather cup containing a stone. The cup is pulled back against two parallel rubber bands. It takes a force of 15 N to stretch either one of these bands 1.0 cm. (a) What is the potential energy stored in the two bands together when a 58-g stone is placed in the cup and pulled back 0.10 m from the equilibrium position?
Final answer:The potential energy stored in the two rubber bands of the slingshot when a 58-g stone is placed in the cup and pulled back 0.10 m from the equilibrium position is 15 J.Explanation:The potential energy stored in the two rubber bands of the slingshot can be calculated using the formula U = ½kx², where k is the force constant of the rubber bands and x is the displacement from the equilibrium position. Given that it takes a force of 15 N to stretch each band by 1.0 cm, we can calculate the force constant as k = F/x = 15 N / 0.01 m = 1500 N/m. Additionally, we know the displacement of the stone as 0.10 m and its mass as58 g.The potential energy stored in one rubber band is U = ½kx² = ½(1500 N/m)(0.10 m)² = 7.5 J. Since there are two rubber bands, the potential energy stored in both bands together is twice that amount, which is15 J.Learn more about Potential energy of rubber bands in a slingshot here:brainly.com/question/14687790#SPJ11...