tutorial sheet 6 solutions

# tutorial sheet 6 solutions - Solution of Tutorial Sheet 6 1...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution of Tutorial Sheet 6 1. We can start from P = Fv. Hence, This leads to an ordinary differential equation (ODE) in the form of: This We can do separation of variables Since the initial condition is v=0 at t=0, we can integrate previous equation This leads to an expression of v as a function of time. 2 2 √ To get x(t) we need to integrate velocity To 22 3 To get a(t) we differentiate v(t) 12 2 If m=1000 kg, P=100 kW, and v =100 km/h=27.7 m/s, we can calculate the time needed from v(t) expression. 2 3.84 . 2. A) The relation of angle and height of mass is: cos The potential energy is: cos Therefore, the change in potential energy is: he Δ cos cos 1 B) Change in potential energy is equal to the kinetic energy gain. Change Δ Δ 0Δ 1 cos C) From B, we get 1 1 cos 2 2 1 cos Hence, the radial acceleration is 2 1 cos For tangential acceleration, we can get it from Newton’s 2nd law easily. eleration, sin D) Utilizing Newton’s Law again Law Mass flies off means the normal force is zero. Hence, we get cos And since the mechanical energy is conserved nserved 1 0 cos 2 2 cos 3 Approximately the angle is 48 degrees. Frictional force in opposite direction of direction of motion 3. A) B) . Spring pulls mass in the opposite direction of direction of mass motion C) Yes, normal force and gravitational force. They do no work since their direction is perpendicular to direction of motion. cos D) The mechanical energy at initial condition will be transformed into mechanical energy at the final condition and used to do work. 1 1 1 00 2 2 2 1 0 2 1 4 2 2 Since negative solution doesn’t have physical reason, we only use one solution (the positive length) 2 4. In the system, mechanical energy is conserved. 1 0 sin 2 2 1.24 2 sin / 1 2 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern