# Tutor profile: Kevin B.

## Questions

### Subject: Physics

What is the velocity just before the ground (x = 0) of a particle that has a mass of 10 kg and is dropped from a height of 5 m? (Assume the particle is at rest when is is dropped)

This problem can be solved using two different methods: 1) acceleration and velocity equations, and 2) work-energy theorems. 1) The particle has an initial velocity v_i = 0 m/s. The particle experiences a constant acceleration of -9.8 m/s^2 due to gravity. The negative is due to the choice of coordinate system given in the problem. The ground is zero, and the plane where we dropped it is x = 5 m. First, we develop an equation to describe the velocity: v(t) = a*t + c1. We don't yet know the time in flight, so we also must determine the position equation: x(t) = a*t^2/2 + c1*t + c2. We know two initial conditions and can now solve this system of equations. v(0) = 0 = c1, and x(0) = 5 = c2. Thus, x(t) = a*t^2/2+5. Now, we want to know the time when x(t) = 0, this is given by solving 0 = a*t^2/2+5. or t = sqrt(-10/a). From before, a = -9.8 m/s^2, so t = 1.01 s. Finally, we can use the equation for velocity to find v(1.01 s) = (-9.8 m/s^2)*(1.01 s) = -9.89 m/s. The negative is imperative due to the choice of coordinate system. 2) The problem lends itself to work-energy principles. The total energy of the system in the initial and final states must remain the same, but the balance of potential and kinetic energy has changed. In the initial state, the potential energy is determined by the height of the particle, which is given by PE = m*g*h_1, where m is the mass of the particle, g is the gravity, and h is the height of the particle from the ground (h_1 = 5 m). In the initial state, the kinetic energy is given by (1/2)*m*v_1^2, but v_1 = 0 m/s. The final state has the same equations for potential and kinetic energy, but the height is now 0 (h_2 = 0), and the velocity is what we want to find (v_2). Thus, the final solution is given by the following equation: m*g*h_1 + (1/2)*m*v_1^2 = m*g*h_2 + (1/2)*m*v_2^2. When we plug in the numbers for our equation, we find m*g*5 + (0) = (0) + (1/2)*m*v_2^2. This results in v_2 = sqrt(g*10) = 9.89 m/s. The energy method does not explicitly provide a sign to the velocity; however, based on the context of the problem, we can determine that the velocity is negative, and should be reported as v = -9.89 m/s.

### Subject: Mechanical Engineering

If a 2300 kg vehicle is travelling at 30 kph and hits an obstacle causing it to stop after 1 s what is the force on the the object that was hit? If the object that was hit was a cantilevered cylindrical beam with a Young's modulus of 200 GPa, diameter of 100 cm and length of 2 m, what is deflection caused by this force if it was concentrated at the end of the post?

This two-part question takes both dynamics and mechanics into account. The dynamic portion is determined by using the basic assumption of F=ma. The mass of the vehicle is already in the correct units at 2300 kg. The speed of the vehicle should be converted to m/s for use in the equation. This is done by the following equation: (30 kph)*(1000 m/1km)*(1 h/3600 s) = 8.3 m/s. The acceleration is determined by the time it took the velocity to reach 0 m/s, which is 1 s. The acceleration is then 8.3 m/s / (1 s) = 8.3 m/s^2. The force is then determined by F = (2300 kg)*(8.3 m/s^2) = 19 kN. The second portion of this question is a challenging mechanics problem that makes use of Castigliano's method. Castigliano's method can be solved in general for a number of loading cases, but the deflection at the end of a cantilevered beam is a common solution given by delta = F*L^3/(3EI), where F is the force, L is the length of the beam, E is the Young's modulus of the material, and I is the second moment of area of the cross section. The second moment of area for a circular cross section is a common result that is given by I = (pi/4)*r^4 = (pi/64)*d^4. In this example, E = 200e9 Pa, L = 2 m, I = (pi/4)*(0.1)^4 = 4.9e-6 m^4, and the force is given by the solution to part 1: 19e3 Pa. When we plug this into our equation, the result is delta = ( (19e3 Pa) * (2 m )^3) ) / ( 3 * (200e9 Pa) * (4.9e-6 m^4) ) = 0.052 m.

### Subject: Calculus

If an airplane flies with a velocity that is steadily increasing in a linear fashion, what is the shape of the curve that defines the position of the airplane.

The position of the airplane is defined by a quadratic equation. It will have a concave up shape and be of the form A*x^2+B*x+C, where A is defined by the rate at which the velocity is increasing.

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