A small ball of mass 150 g is placed at a height of 20cm above the ground on an incline of 35°. It is released and allowed to roll down the slope; what will be the ball's speed when it reaches the ground? Assume friction and air resistance can be ignored.
This question is best solved using conservation of energy, since it is stated that friction and drag (non-conservative forces) can be ignored.When the ball is released at the top of the slope, its initial speed is zero. Therefore, at the top of the slope, the ball has no kinetic energy; all its energy is purely potential energy.The equation for potential energy is U=mgh, where m is the mass of the ball, g is acceleration due to the Earth’s gravitational attraction, and h is the height of the ball above the horizontal.When the ball has reached the end of the slope, h is equal to zero, and therefore the potential energy is now zero; all the ball’s energy is purely kinetic energy, where K=1/2mv2 (v is the velocity of the ball) Conservation of energy states that the initial energy of a closed system has to be equal to its final energy. Here, the initial energy is simply its potential energy at the start, so Ei=U=mgh, and the final energy is the kinetic energy, so Ef=K=1/2mv2.Equating the two gives mgh=1/2mv2. Dividing both sides by m yields gh=1/2v2, and multiplying both sides by two and finally taking the square root leads to v=(2gh)1/2.Now, converting the given values into SI units and putting them into this equation gives us a value for v:v=(29.810.2)1/2 = 1.98 m/s
Note: since both m and the angle of incline are given, it might be tempting to try and use equations of motion and force balance to solve this question; however, doing so by conservation of energy is much quicker and easier.
