Investigating Forces & Motion

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INI File | 1998-02-10 | 10KB | 338 lines

[question1] type:2 caption:When there is no force acting on a moving object: correct:It moves at a steady speed in a straight line. wrong1:It slows down and stops. wrong2:It moves in a circle. wrong3:It accelerates. feedback:\ Newton's first law of motion tell us that when there is no net force \ on a moving object, it will continue to move at a steady speed in a \ straight line. [question2] type:2 caption:Newton's first law of motion states that: correct:\ When no forces act, an object remains at rest or moving at a steady \ speed in a straight line. wrong1:When no forces act, moving objects always come to rest. wrong2:A force is needed to keep an object moving. wrong3:When no forces act, an object moves in circles. feedback:\ Newton's first law of motion tells us that when no force is acting, an \ object remains at rest or moving at a steady speed in a straight \ line. [question3] type:3 caption:\ Which of the following equations is a correct form of Newton's second \ law of motion? correct:4g18a wrong1:4g18b wrong2:4g18c wrong3:4g18d feedback:\ According to Newton's second law of motion F = ma. This \ equation can be rearranged as:\ m = F/a [question4] type:2 caption:\ You pull hard on a rope tied to a tree. According to Newton's third \ law of motion, for every action there is an equal and opposite \ reaction. What is the reaction force to your pull on the rope? correct:The pull of the rope on your hands. wrong1:The pull of the rope on the tree. wrong2:The pull of the tree on the rope. wrong3:The push of your feet against the ground. feedback:\ The equal and opposite action and reaction forces always act on \ different objects - that is, the two objects that are applying forces \ to each other. The reaction to your pull on the rope is the rope's \ pull on you. [question5] type:2 caption:What is the correct definition of one newton? correct:1.0 kg x 1.0 m/s2 wrong1:1.0 kg x 1.0 m wrong2:1.0 kg x 1.0 m/s wrong3:1.0 kg x 1.0 s feedback:\ One newton is the force that gives a mass of 1.0 kg an acceleration of \ 1.0 m/s2. Since F = ma:\ \ 1.0 N = 1.0 kg x 1.0 m/s2. [question6] type:2 caption:\ A 2.0 kg mass is accelerating at 5.0 m/s2. What force is \ producing the acceleration? correct:10 N wrong1:2.0 N wrong2:2.5 N wrong3:8.0 N feedback:\ Use the equation:\ F = ma\ \ <center>= 2.0 x 5.0</center>\ <center>= 10 N.</center> [question7] type:2 caption:\ A force of 20 N accelerates a 5.0 kg mass. What is the size of the \ acceleration produced? correct:4.0 m/s2 wrong1:20 m/s2 wrong2:16 m/s2 wrong3:5.0 m/s2 feedback:\ If F = ma, then:\ <img src="sa4q7a" align=center>\ <img src="sa4q7b" align=center>\ <center>= 5.0 m/s2.</center> [question8] type:2 caption:\ A force of 100 N produces an acceleration of 0.1 m/s2 when \ it is applied to a mass. What is the size of the mass? correct:1 000 kg wrong1:0.001 kg wrong2:10 kg wrong3:100 kg feedback:\ If F = ma, then:\ <img src="sa4q8a" align=center>\ <img src="sa4q8b" align=center>\ <center>= 1 000 kg.</center> [question9] type:2 caption:\ Identical forces applied to two objects A and B make object B \ accelerate six times more rapidly than object A. If the mass of object \ A is 3.0 kg, what is the mass of object B? correct:0.5 kg wrong1:1.0 kg wrong2:3.0 kg wrong3:18 kg feedback:\ Since the forces are identical then, using F = ma, we \ find that:\ mass of A x acceleration of A = mass of B x acceleration of B\ Rearranging this equation, we find that:\ <img src="sa4q9a" align=center>\ Therefore,\ <img src="sa4q9b" align=center>\ <img src="sa4q9c" align=center>\ <center>= 0.5 kg.</center> [question10] type:2 caption:\ Object A has twice the mass of object B. When identical forces are \ applied to the two objects, object A accelerates at 10 \ m/s2. What is the acceleration of object B? correct:20 m/s2 wrong1:10 m/s2 wrong2:5.0 m/s2 wrong3:2.0 m/s2 feedback:\ Since the forces are identical then, using F = ma, we \ find that:\ mass of A x acceleration of A = mass of B x acceleration of B\ Rearranging this equation, we find that:\ <img src="sa4q10a" align=center>\ Therefore,\ <img src="sa4q10b" align=center>\ <center>= 10 x 2.0</center>\ <center>= 20 m/s2.</center> [question11] type:1 image:4g16 caption:\ A boat is being towed at a steady speed of 2.0 m/s. If the total air \ and water resistance opposing the boat's motion is 200 N, what is the \ tension force in the rope? correct:200 N wrong1:100 N wrong2:198 N wrong3:400 N feedback:\ The boat's speed is steady. According to Newton's first law of motion, \ this means that the net force on the boat is zero. If the forces \ opposing motion total 200 N, the tension force in the rope must be 200 \ N in the opposite direction to balance this force. [question12] type:2 caption:\ A 0.5 kg ball is falling towards the ground with the acceleration due \ to gravity 9.8 m/s2. What is the size of the force of \ gravity producing this acceleration? correct:4.9 N wrong1:0.5 N wrong2:9.8 N wrong3:10 N feedback:\ Use the equation:\ F = ma\ \ F = 0.50 x 9.8\ <center>= 4.9 N.</center> [question13] type:2 caption:The weight of an object is: correct:The force of gravity on the object. wrong1:The object's mass. wrong2:The object's acceleration due to gravity. wrong3:The object's density. feedback:An object's weight is the force of gravity acting on it. [question14] type:2 caption:\ The acceleration due to gravity on Planet X is 4.9 m/s2. If \ you weigh 520 N on Earth, what would your weight be on Planet X? \ (Acceleration due to gravity on Earth = 9.8 m/s2.) correct:260 N wrong1:zero wrong2:130 N wrong3:520 N feedback:\ The acceleration produced by gravity on Planet X is half that on the \ Earth. This means that the force of gravity on any object on Planet X \ must also be half its value on Earth. The weight of an object on \ Planet X is therefore half that of the same object on Earth. \ Therefore,\ 520 x ½ = 260 N. [question15] type:2 caption:\ The acceleration due to gravity on Planet X is 2.4 m/s2. If \ your mass is 52 kg on Earth, what would your mass be on Planet X? \ (Acceleration due to gravity on Earth = 9.8 m/s2.) correct:52 kg wrong1:zero wrong2:12 kg wrong3:16 kg feedback:\ Mass is a fixed property of an object, and does not change when you \ move from one planet to another. So, although you only weigh a quarter \ as much on Planet X as you do on Earth, your mass remains the same - \ that is, 52 kg. [question16] type:2 caption:\ A space-walking astronaut manoeuvres by firing small gas jets from his \ backpack. Each jet produces a force of 4.0 N. If firing a single jet \ produces an acceleration of 0.02 m/s2, what is the combined \ mass of the astronaut, space suit and backpack? correct:200 kg wrong1:20 kg wrong2:80 kg wrong3:800 kg feedback:\ If F = ma, then:\ <img src="sa4q16a" align=center>\ <img src="sa4q16b" align=center>\ <center>= 200 kg.</center> [question17] type:2 caption:\ The rockets on a spacecraft produce a thrust of 40 000 N. The engines \ are fired for 50 s, accelerating the craft in a straight line through \ space. If the total mass of the spacecraft is 8 000 kg, by how much \ does its velocity change? correct:250 m/s wrong1:5.0 m/s wrong2:20 m/s wrong3:200 m/s feedback:\ If F = ma, then:\ <img src="sa4q17a" align=center>\ <img src="sa4q17b" align=center>\ <center>= 5.0 m/s2</center>\ Also\ <img src="sa4q17c" align=center>\ Therefore,\ velocity change = acceleration x time\ <center>= 5.0 x 50</center>\ <center>= 250 m/s.</center> [question18] type:2 caption:\ A tennis ball with a mass of 0.05 kg is in contact with a racket for \ 0.005 s. During this time, its velocity changes by 30 m/s. What was \ the average force exerted by the racket on the ball? correct:300 N wrong1:15 N wrong2:30 N wrong3:150 N feedback:\ Use the equation:\ <img src="sa4q17c" align=center>\ <img src="sa4q18a" align=center>\ <center>= 6 000 m/s2</center>\ Also\ F = ma\ \ <center>= 0.05 x 6 000</center>\ <center>= 300 N.</center> [question19] type:2 caption:\ A racing car can accelerate from rest to 50 m/s in 5.0 s. If the \ average force producing acceleration is 5 000 N, what is the mass of \ the car? correct:500 kg wrong1:100 kg wrong2:1 000 kg wrong3:5 000 kg feedback:\ Use the equation:\ <img src="sa4q17c" align=center>\ <img src="sa4q19a" align=center>\ <center>= 10 m/s2</center>\ If F = ma, then:\ <img src="sa4q16a" align=center>\ <center>= 500 kg.</center> [question20] type:2 caption:\ A footballer kicks a ball with an average force of 1 000 N. The mass \ of the ball is 0.4 kg. If the footballer's boot is in contact with the \ ball for 0.008 s, how fast is the ball travelling when it leaves his \ boot? correct:20 m/s wrong1:10 m/s wrong2:40 m/s wrong3:80 m/s feedback:\ If F = ma, then:\ <img src="sa4q17a" align=center>\ <img src="sa4q20a" align=center>\ <center>= 2 500 m/s2</center>\ Also\ <img src="sa4q17c" align=center>\ Therefore,\ velocity change = acceleration x time\ <center>= 2 500 x 0.008</center>\ <center>= 20 m/s.</center>