ch6

Ch 6. Work Problems---Quiz 7: Questions? email nalexander@igc.org
22	24	26	31	36	45	73	81	82
Helpers!! Click this link then click the links "Practice Questions" and "Practice Problems" in the left hand column.
Final Exam = Monday , December 18
Click here for Real Test 1, Sp 2005. Click here for Real Test 1, Sp 2003.
SP 1. Redo #26 with with friction coefficient u_k = 0.05, using Ch. 6 and Ch. 7 methods.
SP 2 Redo #36 (b) with with friction coefficient u_k = 0.05,, using Ch. 6 and Ch. 7 methods. Note that part (a) would remain the same since it is only asking for the work of the spring...
22. Wnet = W_N + Wg + Ws + W_f+ W_F = KE_f- KE_i . W_F = Fd. W_N = Wg = Ws =W_f = 0 . (a) KE_i = 0 since Vi = 0. Set Fd = KE_f. Find V_f. (b) W_f = -fd where f = µN. What is the normal force magnitude N? W_f+ W_F = KE_f- KE_i . KE_i = 0 since Vi = 0. Find V_f.
24. (a) High to low means Wg = +mgh. (b) Wg = +mgh = KE_f- KE_i . KE_i = 0 since Vi = 0. Set mgh = KE_f. Find V_f.
26. Wnet = W_N + Wg + Ws + W_f+ W_F = KE_f- KE_i W_N = W_F =Ws =W_f = 0 . High to low means Wg = +mgh. h = L sin 36.9, where L = 0.750 m. +mgh = KE_f- KE_i . KE_i = 0 since Vi = 0. Find V_f.
31. (a) W_F = KE_f- KE_i . KE_i = 0 since Vi = 0. Note that the force of magnitude F is not constant, so you have to integrate to get the answer. From 0 to 8 m, W_F = area under the curve of the triangle of height 10 N and base 8 m. Find V_f. (b) W_F = KE_f- KE_i . KE_i = 0 since Vi = 0. You have to integrate to get the answer. From 0 to 12 m, W_F = area under the curve of the triangle of height 10 N and base 12 m. W_F = KE_f- KE_i . KE_i = 0 since Vi = 0. Find V_f.
36. (a) Total Work = Ws, where Ws = work of spring, which is positive. Ws = ½ kx₁² - ½ kx₂² . Note: x₂ = 0. (b) Total Work = change in KE = ½ mv₂² - ½ m₂v₁² . Find the second speed. The first speed = 0.
45. Power = Fv
73. (a) Total Work = change in KE = ½ mv₂² - ½ m₂v₁² . The initial and final speed are given. (b) Total Work = W_F + Wg, where W_F = work by your force F on the pedals. Wg = -mgh. Solve for W_F .
81. (a) Total Work = Ws, where Ws = work of spring, which is negative. Ws = ½ kx₁² - ½ kx₂² . Note: x₁ = 0. Total Work = change in KE = ½ mv₂² - ½ m₂v₁² . v₁ is given and v₂ = 0. Find the value x₂. (b) Total Work = Ws = change in KE = ½ mv₂² - ½ m₂v₁² . Find the first speed. The second speed = 0.
82. W_Tot = ½ m₁v² + ½ m₂v² W_f + Wg = ½ m₁v² + ½ m₂v² -u_kNh + m₂gh = ½ m₁v² + ½ m₂v² m₁ = 8.00 kg and m₂ = 6.00 kg What is N ? Hint: It is the N for m₁. Solve for v. Note that the work by tension for m₁ is T·h and the work by tension for m₂ is -T^·h. So these works cancel. Think about it !!
If you think there are errors in the hints, email me and I will send out an alert on the correction if I agree there is a mistake to be fixed. peace.