part c - fix the false (write t or f; if false, correct the idea in a few words)\n1. a horizontal push that…

part c - fix the false (write t or f; if false, correct the idea in a few words)\n1. a horizontal push that doesnt move the box still does work. __correction:__\n2. 1 watt is the same as 1 n·m. __correction:__\n3. kinetic energy depends only on mass. __correction:__\n4. lifting the same object to the same height does more work if you lift it faster. __correction:__\npart d - calculations (show all steps; include units)\n1. a student pulls a cart with a constant horizontal force of 35 n for 3.2 m in the same direction. how much work is done on the cart?\n2. a motor does 4,500 j of work in 90 s. what is its power output?\n3. a drill operates at a constant power of 180 w for 50 s. how much work (energy) does it do in that time?\n4. an 18.0 kg backpack is lifted to a shelf 1.80 m high. find the backpacks gravitational pe relative to the floor.\n5. a 1.50 kg textbook slides at 8.0 m/s across a smooth table. find its kinetic energy.\n6. a 0.45 kg soccer ball moves at 15.0 m/s. calculate its ke.\n7. a rope pulls a sled 7.5 m across level ground with a constant horizontal force of 42 n in the same direction. calculate the work done by the rope.\n8. a winch delivers 3,600 j of work at an average power of 120 w. how long did it run (time t)?\nformula bank (for review)\n- work: (w = f·d) (force and displacement along the same line)\n- power: (p = ) so (w = p·t)

part c - fix the false (write t or f; if false, correct the idea in a few words)\n1. a horizontal push that doesnt move the box still does work. __correction:__\n2. 1 watt is the same as 1 n·m. __correction:__\n3. kinetic energy depends only on mass. __correction:__\n4. lifting the same object to the same height does more work if you lift it faster. __correction:__\npart d - calculations (show all steps; include units)\n1. a student pulls a cart with a constant horizontal force of 35 n for 3.2 m in the same direction. how much work is done on the cart?\n2. a motor does 4,500 j of work in 90 s. what is its power output?\n3. a drill operates at a constant power of 180 w for 50 s. how much work (energy) does it do in that time?\n4. an 18.0 kg backpack is lifted to a shelf 1.80 m high. find the backpacks gravitational pe relative to the floor.\n5. a 1.50 kg textbook slides at 8.0 m/s across a smooth table. find its kinetic energy.\n6. a 0.45 kg soccer ball moves at 15.0 m/s. calculate its ke.\n7. a rope pulls a sled 7.5 m across level ground with a constant horizontal force of 42 n in the same direction. calculate the work done by the rope.\n8. a winch delivers 3,600 j of work at an average power of 120 w. how long did it run (time t)?\nformula bank (for review)\n- work: (w = f·d) (force and displacement along the same line)\n- power: (p = ) so (w = p·t)

Answer

Explanation:

Part C - Fix the False

1.

  • Explanation: Work is defined as $W = Fd\cos\theta$, where $\theta$ is the angle between force and displacement. If $d = 0$, $W=0$.
  • Answer: F, A horizontal push that doesn't move the box does no work.

2.

  • Explanation: 1 Watt is power, and 1 N·m is work. 1 W = 1 J/s, 1 J=1 N·m.
  • Answer: F, 1 Watt is the same as 1 J/s.

3.

  • Explanation: Kinetic energy is $K=\frac{1}{2}mv^{2}$, depends on mass $m$ and velocity $v$.
  • Answer: F, Kinetic energy depends on mass and velocity.

4.

  • Explanation: Work done in lifting an object to a height $h$ is $W = mgh$, which only depends on the weight $mg$ and height $h$, not the time.
  • Answer: F, Lifting the same object to the same height does the same work regardless of speed.

Part D - Calculations

1.

  • Explanation: Use the work - formula $W = Fd$. Given $F = 35$ N and $d=3.2$ m.
  • Expression: $W=Fd=(35\ N)\times(3.2\ m)$
  • Answer: $W = 112$ J

2.

  • Explanation: Use the power formula $P=\frac{W}{t}$. Given $W = 4500$ J and $t = 90$ s.
  • Expression: $P=\frac{W}{t}=\frac{4500\ J}{90\ s}$
  • Answer: $P = 50$ W

3.

  • Explanation: Use the formula $W = Pt$. Given $P = 180$ W and $t = 50$ s.
  • Expression: $W=Pt=(180\ W)\times(50\ s)$
  • Answer: $W = 9000$ J

4.

  • Explanation: Use the gravitational - potential energy formula $PE=mgh$. Given $m = 18.0$ kg, $g = 9.8\ m/s^{2}$, and $h = 1.80$ m.
  • Expression: $PE=mgh=(18.0\ kg)\times(9.8\ m/s^{2})\times(1.80\ m)$
  • Answer: $PE\approx317.5$ J

5.

  • Explanation: Use the kinetic - energy formula $K=\frac{1}{2}mv^{2}$. Given $m = 1.50$ kg and $v = 8.0$ m/s.
  • Expression: $K=\frac{1}{2}mv^{2}=\frac{1}{2}\times(1.50\ kg)\times(8.0\ m/s)^{2}$
  • Answer: $K = 48$ J

6.

  • Explanation: Use the kinetic - energy formula $K=\frac{1}{2}mv^{2}$. Given $m = 0.45$ kg and $v = 15.0$ m/s.
  • Expression: $K=\frac{1}{2}mv^{2}=\frac{1}{2}\times(0.45\ kg)\times(15.0\ m/s)^{2}$
  • Answer: $K\approx50.6$ J

7.

  • Explanation: Use the work formula $W = Fd$. Given $F = 42$ N and $d = 7.5$ m.
  • Expression: $W=Fd=(42\ N)\times(7.5\ m)$
  • Answer: $W = 315$ J

8.

  • Explanation: Use the power formula $P=\frac{W}{t}$, then $t=\frac{W}{P}$. Given $W = 3600$ J and $P = 120$ W.
  • Expression: $t=\frac{W}{P}=\frac{3600\ J}{120\ W}$
  • Answer: $t = 30$ s