# Tutor profile: Austin B.

## Questions

### Subject: Mechanical Engineering

Why might writing skills be valuable to a technically trained Mechanical Engineer?

Being able to effectively communicate through writing is an extremely valuable asset for a Mechanical Engineer. It is important for non MechE's to be able to understand values you have determined/reasoning why you are making specific recommendations.

### Subject: Calculus

What is the area under the curve f(x) = 4x^3 + 3x +7 from x = -0.75 to x = -0.10?

Integrate f(x) dx from -0.75 to -0.10. Take the antiderivative of f(x): F(x) = x^4 + 1.5x^2 +7x Area = [(-0.10)^4 + (1.5)*(-0.10)^2 + (7)*(-0.10)] - [(-0.75)^4 + (1.5)*(-0.75)^2 + (7)*(-0.75)] Are = 3.405

### Subject: Physics

A ball with a mass of 2 kg is held in the air 5 meters above the ground and is dropped. Determine both the potential and kinetic energy for these three instances: (a) right before the ball is dropped, (b) when the ball is 3 meters above the ground, and (c) the moment the ball hits the ground.

Potential energy = mgh, where m is mass, g is the gravitational constant, and h is the height above the ground. Kinetic energy = (1/2)m(v^2), where m is mass and v is the velocity of the object. (a) PE = (2 kg)*(9.8 m/s^2)*(5m) = 98 J, KE = 0 because v = 0 right before the ball is dropped. (b) PE = (2 kg)*(9.8 m/s^2)*(3m) = 58.8 J, KE = (1/2)*(2 kg)*v^2 ... we need to solve for v^2 here. We learned from kinematic equations that (v_final)^2 = (v_initial)^2 + 2*a*d. v_final is the v we want, v_initial is 0 because the ball's velocity is 0 before it is dropped, a is acceleration which is g in this problem, and d is the distance traveled from 5 m to 3 m (which is 2 m). Now we get v^2 = (0^2)+(2)*(9.8)(2) = 39.2 m^2/s^2. So now KE = (1/2)*(2 kg)*(39.2 m^2/s^2) = 39.2 J (c) PE = 0 because h = 0. We can solve for KE using the same method as is part (b); however, we can apply conservation of energy to determine that KE = 98 J. Notice in part (b) that 58.8 J + 39.2 J = 98 J.

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