This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 106 – TEST02 – TSOI – (58160) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 4 kg crate slides down a ramp at a loading dock. The ramp is 3 m in length and inclined at an angle of 34 ◦ , as shown in the figure. The crate starts from rest at the top, experi ences a constant frictional force of magnitude 5 N, and continues to move a short distance on the flat floor. The acceleration of gravity is 9 . 8 m / s 2 . y s θ v = 0 v i f Use energy methods to determine the speed of the crate when it reaches the bottom of the ramp. 1. 5.45218 2. 4.3879 3. 4.65663 4. 3.20033 5. 5.03791 6. 3.9715 7. 2.89593 8. 3.03786 9. 2.77844 10. 4.14136 Correct answer: 5 . 03791 m / s. Explanation: Since v i = 0, the initial kinetic energy is zero. If the y coordinate is measured from the bottom of the ramp, then y i = s sin θ = 1 . 67758 m . Therefore, the total mechanical energy of the system at the top is all potential energy: U i = mg y i = 65 . 7611 J . When the crate reaches the bottom, the po tential energy is zero because the elevation to the crate is y f = 0. Therefore, the total me chanical energy at the bottom is all kinetic energy: K f = 1 2 mv 2 f . However, we cannot say that U i = K f in this case, because there is an external nonconser vative force that removes mechanical energy from the system: the force of friction. In this case, W ext = fs , where s is the displace ment along the ramp. (Remember that the forces normal to the ramp do not do work on the crate because they are perpendicular to the displacement.) With f = 5 N and s = 3 m , we obtain W ext = f s = 15 J . This says that some mechanical energy is lost because of the presence of the retarding fric tional force. So, we have ∆ K + ∆ U = W ext , or 1 2 mv 2 f mg y i = f s, which gives us v f = radicalBigg 2 s parenleftbigg g sin θ f m parenrightbigg = radicalBigg 2(3 m) bracketleftbigg (9 . 8 m / s 2 ) sin 34 ◦ 5 N 4 kg bracketrightbigg = 5 . 03791 m / s . 002 (part 1 of 2) 10.0 points A 3 . 06 kg block is placed on top of a 10 . 7 kg block. A horizontal force of F = 54 . 8 N is applied to the 10 . 7 kg block, and the 3 . 06 kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0 . 159. There is friction both between the masses and between the 10 . 7 kg block and the ground. The acceleration of gravity is 9 . 8 m / s 2 . Version 106 – TEST02 – TSOI – (58160) 2 3 . 06 kg 10 . 7 kg μ = 0 . 159 μ = 0 . 159 F T Determine the tension T in the string. 1. 3.68568 2. 2.1462 3. 3.62392 4. 5.12785 5. 4.98095 6. 7.47848 7. 4.76809 8. 6.77386 9. 4.49261 10. 6.4974 Correct answer: 4 . 76809 N....
View
Full Document
 Fall '10
 TSIO
 Energy, Force, Friction, Mass, Potential Energy, Correct Answer

Click to edit the document details