Cindy is changing the tire on her car on a steep hill, 27.6 m high. she trips and drops the 8.7 kg spare tire, which rolls down the hill with an initial speed of 2.37 m/s. what is the speed of the tire at the top of the next hill, which is 7.9 m high?

Initially, the tire has kinetic energy and potential energy. If we assume 0 potential energy at 7.9 m, then the initial energy is 
E = mgh + ½mv² = 8.7kg * 9.8m/s² * (27.6 – 7.9)m + ½ * 8.7kg * (2.37m/s)² = 1,690 J 
With on no account of energy loss, this will be the energy at the topmost of the 7.9m hill, which we distinct as having 0 potential energy. Then:
E = 1690 J = ½mv² = ½ * 8.7kg * v² v = 41 m/s

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge