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Transcript
Partial Credit - Choose 3 out of 4 problems. (If you do all four problems I take the best 3 scores) 1. In the overhead view of the Figure, five forces pull on a box of mass m = 4.0 kg. The force magnitudes are F1 = 11 N, F2 = 17 N, F3 = 3.0 N, F4 = 14 N, and F5 = 5.0 N, and angle q4 is 30°. Find the box's acceleration (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the x axis. (2points each) With SI units understood, the net force on the box is r Fnet = ( 3.0 + 14 cos 30° − 11) ˆi + (14 sin30° +5.0 − 17 ) ˆj r which yields Fnet = (4.1 N) ˆi − ( 5.0 N) ˆj . (a) Newton’s second law applied to the m = 4.0 kg box leads to r Fnet r a = = (1.0 m/s 2 )iˆ − (1.3 m/s 2 )ˆj . m r 2 2 2 (b) The magnitude of a is a = 1.0 + ( −1.3) = 1.6 m s . (c) Its angle is tan–1 (–1.3/1.0) = –50° (that is, 50° measured clockwise from the rightward axis). 2. The figure shows a kimchi container of mass m1 = 3.0 kg connected to a block of mass m2 by a cord looped around a frictionless pulley. The cord and pulley have negligible mass. When the container is released from rest, it accelerates at 1.0 m/s2 across the horizontal frictionless surface. What are (a) the tension in the cord (4 Points) and (b) mass m2? (2 Points) Using the usual coordinate system (right = +x and up = +y) for both blocks has the important consequence that for the m1 = 3.0 kg block to have a positive acceleration (a > 0), block m2 must have a negative acceleration of the same magnitude (–a). Thus, applying Newton’s second law to the two blocks, we have T = m1a = (3.0 kg) (1.0 m/s 2 ) T − m2 g = m2 ( − 1.0 m/s 2 ) along x − axis along y − axis. (a) The first equation yields the tension T = 3.0 N. (b) The second equation yields the mass m2 = 3.0/8.8 = 0.34 kg. 3. During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. The Figure shows a cross section of Mt. Fuji, in Japan. (a) At what initial speed would a bomb have to be ejected, at angle q0 = 35° to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical distance h = 3.30 km and horizontal distance d = 9.40 km? Ignore the effects of air on the bomb's travel. (4 Points) (b) What would be the time of flight (2 Points) (a) By rearranging Eq. 4-25, we obtain the initial speed: v0 = x cosθ 0 g 2( x tan θ 0 − y ) which yields v0 = 255.5 ≈ 2.6 × 102 m/s for x = 9400 m, y = –3300 m, and θ0 = 35°. (b) From Eq. 4-21, we obtain the time of flight: t= 9400 x = = 45 s. . cos 35° v0 cosθ 0 2555 4. A proton moves along the x axis according to the equation x = 50t + 10t2, where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3.0 s of its motion, (b) the instantaneous velocity of the proton at t = 3.0 s, and (c) the instantaneous acceleration of the proton at t = 3.0 s. (2 points each) In this solution, we make use of the notation x(t) for the value of x at a particular t. Thus, x(t) = 50t + 10t2 with SI units (meters and seconds) understood. (a) The average velocity during the first 3 s is given by v avg = x (3) − x (0) (50)(3) + (10)(3) 2 − 0 = = 80 m / s. Δt 3 (b) The instantaneous velocity at time t is given by v = dx/dt = 50 + 20t, in SI units. At t = 3.0 s, v = 50 + (20)(3.0) = 110 m/s. (c) The instantaneous acceleration at time t is given by a = dv/dt = 20 m/s2. It is constant, so the acceleration at any time is 20 m/s2.