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Chapter 4 Homework 9. A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with magnitude 48.0 N to the box and produces an acceleration of magnitude 3.00 m/s 2 , what is the mass of the box? = Fx max . = . 48 0 N m3 00ms2 = m 16 kg 10. A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) What is the mass of the block of ice? - = + x x0 v0xt 12 axt2 . = . + . 11 0 m 0ms5 00 s 12 ax5 00 s2 = . ax 0 880ms2 = Fx max . = . 80 0 N m0 880ms2 = . m 90 9 kg (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s? = + vx v0x axt = + . . 0ms 0 880ms25 00 s = . / 4 40 m s- = + x x0 v0xt 12 axt2 = . / . + . 4 40 m s5 00 s 12 0ms25 00 s2 = . 22 0 m 12. A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 140 N. (a) What acceleration is produced? = Fx max = . 140 N 32 5 kgax = . ax 4 31ms2 (b) How far does the crate travel in 10.0 s? Page | 1 2.1.11- = + x x0 v0xt 12 axt2 = / . + . . 0 m s10 0 s 12 4 31 ms210 0 s2 = 215 m (c) What is its speed at the end of 10.0 s? = + vx v0x axt = + . . 0ms 4 31ms210 0 s = . / 43 1 m s 16. An electron (mass = 9.11 x 10-31 kg) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 cm away. It reaches the grid with a speed of 3.00 x 10 6 m/s. If the accelerating force is constant, compute (a) the acceleration = +- vx 2 v0x 2 2axx x0 . = + . 3 00 106ms2 0ms2 2ax0 0180 m = . ax 2 50 1014ms2 (b) the time to reach the grid = + vx v0x axt . = + . 3 00 106ms 0ms 2 50 1014ms2 t = . - t 1 20 10 8 s (c) the net force, in newtons = Fx max = . - . 9 11 10 31 kg2 50 1014ms2 = . - 2 28 10 16 N 17. Superman throws a 2400-N boulder at an adversary. What horizontal force must Superman apply to the boulder to give it a horizontal acceleration of 12.0 m/s 2 ? = w mg = . 2400 N m9 80ms2 = m 245 kg = Fx max = . 245 kg12 0ms2 = 2940 N 18. A bowling ball weighs 71.2 N (16.0 Ib). The bowler applies a horizontal force of 160 N (36.0 lb) to the ball. What is the magnitude of the horizontal acceleration of the ball? = w mg Page | 2 2.1.11 . = . 71 2 N m9 80ms2 = . m 7 27 kg = Fx max = . 160 N 7 27 kgax = . / ax 22 0 m s2 19. At the surface of Jupiter's moon Io, the acceleration due to gravity is g = 1.81 m/s 2 . A watermelon weighs 44.0 N at the surface of the earth. (a) What is the watermelon's mass on the earth's surface? = w mg . = . 44 0 N m9 80ms2 = . m 4 49 kg (b) What are its mass and weight on the surface of Io?... View Full Document