This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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You're on a roll — Step 1: Answer Question 15 (a). To show that the potential energy gained by a body of mass m lifted vertically through a height h is E = mgh: When a body is lifted, work is done against the force of gravity. The force of gravity (weight) acting on the body is given by: F = mg Where m is the mass of the body and g is the gravitational field strength. The work done (W) in lifting the body through a vertical height h is the product of the force and the distance moved in the direction of the force: W = F × h Substituting the force F = mg: W = mg × h This work done against gravity is stored in the body as gravitational potential energy (E). Therefore, the potential energy gained by the body is: E = mgh This completes the derivation. Step 2: Answer Question 15 (b) i. Given: Mass of the ball m = 350 kg Energy to be delivered E = 10 kJ = 10 × 1000 J = 10000 J Gravitational field strength g = 9.8 m s^-2 The energy delivered to the wall comes from the gravitational potential energy of the ball. E = mgh To find the height h: h = (E)/(mg) h = 10000 J350 kg × 9.8 m s^-2 h = (10000)/(3430) m h ≈ 2.915 m The ball must be raised to a height of 2.92 m. Step 3: Answer Question 15 (b) ii. Assuming the ball does not lose any energy to the surroundings, the potential energy at its maximum height is converted entirely into kinetic energy at the moment of impact. The kinetic energy (KE) at impact is equal to the energy delivered: KE = E = 10000 J The formula for kinetic energy is: KE = (1)/(2)mv^2 Where v is the velocity at impact. 10000 J = (1)/(2) × 350 kg × v^2 10000 = 175 v^2 v^2 = (10000)/(175) v^2 ≈ 57.1428 v = sqrt(57.1428) v ≈ 7.559 m s^-1 The velocity at the moment of impact is 7.56 m s^-1. Got more? Send 'em!