Question 01. A. Incorrect! This is not Newton s second law.

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1 College Physics - Problem Drill 06: Newton s Laws of Motion Question No. 1 of Which of the options best describes the statement: Every object continues in a state of rest or uniform motion in a straight line, unless it acted upon by an outside force. Question 01 (A) Newton s second law of motion, the law of inertia. (B) Newton s first law of motion. The tendency to keep moving is called inertia. (C) For every action there is an equal and opposite reaction. (D) Newton s first law of motion, the law of action and reaction. (E) Newton s third law of motion, the law of inertia. This is not Newton s second law. B. Correct! This is a statement of the Newton s first law of motion, which is often called the law of inertia. For every action there is an equal and opposite reaction is a statement of Newton s third law. The statement does describe Newton s first law of motion, but it is not about action and reaction. Newton s third law is not the law of inertia. Newton s laws are: Newton s First Law of Motion (the law of inertia): Every object continues in its state of rest, or uniform motion in a straight line, unless acted upon by an outside force. Newton s Second Law of Motion: F net = ma Newton s Third Law of Motion: For every force there is an equal and opposite force. For every action there is an equal and opposite reaction. The correct answer is (B).

2 Question No. 2 of The acceleration due to gravity on Jupiter is approximately 26 m/s 2. Which of these statements best describes your mass and weight on Jupiter? Question 02 (A) Your mass and weight will both be the same as they are on Earth. (B) Your mass will be the same as on earth and weight will be 0.37 times that on Earth. (C) Your mass will be 2.7 times that on earth and weight will be the same. (D) Your mass will be the same and weight will be 2.7 times that on Earth. (E) Your mass and weight will both be 2.7 times what they are on Earth. Incorrect! Remember that weight depends on acceleration due to gravity. Weight is proportional to acceleration due to gravity. Mass is a measure of the amount of material, you have the same number of atoms and molecules on Jupiter as on Earth. D. Correct! Mass is a measure of the amount of material, you have the same number of atoms and molecules on Jupiter as on Earth. Weight is a measure of the force due to gravity, so for a higher acceleration due to gravity on Jupiter you will have a higher weight. Mass is a measure of the amount of material, you have the same number of atoms and molecules on Jupiter as on Earth. Mass is a measure of the amount of material, you have the same number of atoms and molecules on Jupiter as on Earth. The formula for weight is W=mg, where m is mass and g is the acceleration due to gravity on that planet. Weight on Jupiter: W J = mg J Weight on Earth: W E = mg E W mg 2 J J 26 m/s = = = 2.7 Mass is the same on both planets so the m cancels W mg 2 E h 9.8 m/s out. W J = 2.7W E, The correct answer is (D).

3 Question No. 3 of Which of these is a true statement? Question 03 (A) F net = ma, units of force are, kg m/s or Newton (B) F net = m/a, units of force are, kg m/s 2 or Newton (C) F net = ma, units of force are, kg m/s 2 or Watt (D) F net = ma, units of force are kg m/s 2 or Newton (E) F net = ma, units of force kg m/s 2 or Watt Check the units. Force is not inversely proportional to acceleration. The unit of force is not the Watt. D. Correct! Newton s second law of motion is F net = ma so the units are kg m/s 2 which is also called a Newton. kg m/s 2 is not the same as a Watt. Newton s second law of motion is F net = ma so the units are kg m/s 2 which is also called a Newton. The correct answer is (D).

4 Question No. 4 of A 1200 kg car travelling at 100 km/h is brought to rest in 8.0 s, what is the average net force required to stop the car? Question 04 (A) N (B) N (C) -4.2 N (D) N (E) N To find the average net force, first calculate the acceleration, and then use Newton s second law of motion. Check the calculation. Check that you are using the correct calculation for acceleration. Check your conversion from km to m. Check your conversion from hr to s. E. Correct! To find the average net force, first calculate the acceleration, and then use Newton s second law of motion. Known: Initial velocity, v i = 100 km/h Final velocity, v f = 0 m/s (stopped) Time, t = 8.0 s Mass, m =1200 kg Unknown: Average Net Force =? N Define: Output: Convert 100 km/h to m/s; 1 km= 1000 m; 1h =3600 s f i Calculate acceleration : a = Calculate Force: F net = ma v v km 1000 m 1 h 100 km/h = 100 = 27.8 m/s h 1 km 3600 s t m/s f 2 a = = 3.5 m/s 8.0 s F net = 1200 kg x 3.5 m/s 2 =4200 kg m/s 2 = N Substantiate: Units are correct, sig figs are correct, Magnitude is correct, direction is correct. Note the negative sign indicates that the force is in the direction that opposes the motion of the car. The correct answer is (E).

5 Question No. 5 of A 6.1 kg store sign is hanging by a rope off a pole, the pole is attached to the wall of the store by a cable, the angle between the cable and wall is 50. If the sign remains hanging stationary in place, which of these best describes the situation? Question 05 (A) The system is in equilibrium, horizontal components of force balance each other out, the vertical components of force balance each other out, and the cable is under tension. (B) The system is in equilibrium, horizontal components of force balance out the vertical components of force, the cable is under tension. (C) The system is in equilibrium, the vertical components of force, balance out each other, the cable is under tension. (D) The system is in equilibrium, horizontal components of force balance each other out, and the cable is under tension. (E) The system is in equilibrium, horizontal components of force balance each other out, the vertical components of force balance each other out, and there is no tension in the cable. A. Correct! In order for the sign to be held in place the system must be in equilibrium and for this to be true the cable must be in tension. The horizontal forces do not have to be the same as the vertical forces for the system to be in equilibrium. This statement is correct but it is not the best description. This statement is correct but it is not the best description. The cable must be under tension. The system described is in static equilibrium, for the sign to remain hanging stationary in place the net force must be zero. The system is in equilibrium, horizontal components of force balance each other out, the vertical components of force balance each other out, and the cable is under tension. The correct answer is (A).

6 Question No. 6 of kg store sign is hanging by a rope off a pole, the pole is attached to the wall of the store by a cable, the angle between the cable and wall is 50. If the sign remains hanging stationary in place, calculate the tension in the cable, you can ignore the mass of the rod. Question 06 (A) 93 N (B) 9.5 N (C) 78 N (D) 60 N (E) 50 N A. Correct! First find the weight of the sign and then use the formula for Cos θ to find the tension in the cable, where the cable forms the hypotenuse of the cable and the adjacent is equal in magnitude to the weight of the sign. You need to find the weight of the sign. Check that you used the correct trig function. It may help to draw a diagram. This is the weight of the sign, not the tension in the cable. Check that you used the correct trig function. It may help to draw a diagram. Known: Mass of sign, m=6.1 kg Angle of cable to wall = 50 Unknown: Tension in cable F T =? N F V Vertical forces 50 Define: Calculate weight of sign to find F v F H =mg Using sign convention of down is negative. Acceleration due to gravity, g=9.8 m/s 2 The vertical force from the cable Sign must balance 61 the weight of the sign. Calculate tension in cable using: adj F F v v cos θ = =, F = hyp F T cos θ T F T F V Output: F V = 6.1 kg x 9.8 m/s 2 = 60 N 60 N 60 N F = = = T cos N Substantiate: Units are correct, sig figs are correct, Magnitude is reasonable. The correct answer is (A).

7 Question No. 7 of Consider a 100 N weight hanging from two cables attached to walls as shown. Calculate the tension/force in the second cable, T2. Question 07 (A) 115N (B) 100N (C) 145N (D) 200N (E) 1127 N 30 o T2 T1 100N A. Correct! Consider the tension in T2 as horizontal and vertical components. Consider the right triangle formed where the vertical component is 100N to balance the hanging weight. Find the hypotenuse which is the force on T2. cos (30) = 100N/T2 Solving for T2 gives 115N. Consider the tension in T2 as horizontal and vertical components. Only the vertical component of T2 would equal 100N. Consider the tension in T2 as horizontal and vertical components. The horizontal component of T2 must be equal to T1. The vertical component of T2 must be equal to the hanging weight. All of these components add together to give a net force of 0, static equilibrium. To find the tension in T2, consider the right triangle its components form. We know the vertical component is 100N since that must equal the hanging weight. Consider the tension in T2 as horizontal and vertical components. Only the vertical component of T2 would equal 100N. Known: Weight =100 kg Angle of cable = 30.0 Unknown: Tension in cable T2=? N 30 o T2 Define: equilibrium The weight is in static F net = 0 N The vertical component of the tension, T2, is equal to the weight. 100 N T1 Calculate T2 using: adj W W cos θ = =, T2 = hyp T2 cos θ Output: 100 N 100 N T2 = = = 115 N cos Substantiate: Units are correct, sig figs are correct and magnitude is reasonable. The correct answer is (A).

8 Question No. 8 of A 10.0 kg sled is pushed at a constant velocity of 4.0 m/s along a snowy surface. A frictional force of 10.0 N acts on it. What is the net force acting on the sled? Question 08 (A) 196 N (B) 108 N (C) 98 N (D) 88 N (E) 0 N Consider all forces acting on the sled to obtain the net force. Be sure to consider their direction too. Consider all forces acting on the sled to obtain the net force. Be sure to consider their direction too. This would be the weight of the sled, but that s only one force acting on it. Consider all others to get the net force. Consider all forces acting on the sled to obtain the net force. Be sure to consider their direction too. E. Correct! Since the sled is moving at a constant velocity, the acceleration of zero. Since it has zero acceleration, the net force acing on the sled must also be zero. The frictional force is exactly balanced by the applied push, giving a net force of zero. Since the sled is moving at a constant velocity, the acceleration of zero. Since it has zero acceleration, the net force acing on the sled must also be zero. The frictional force is exactly balanced by the applied push, giving a net force of zero. The weight too, is balanced by another force, the normal force. In all direction, the forces acting on the object yield a resultant of zero. The correct answer is (E).

9 Question No. 9 of A box slides down a frictionless inclined plane. It makes an angle of 40 degrees with the horizontal. What is the acceleration of the box as it slides down the incline? Question 09 (A) 0 m/s/s (B) 6.3 m/s/s (C) 7.5 m/s/s (D) 9.8 m/s/s (E) Insufficient data. This would be correct if enough friction could hold the box in place, or if it slid at a constant velocity. In this case, it is a frictionless surface. B. Correct! Find the component of the weight that pulls the box down the incline. This component is the parallel force. This will also be used as the net force. F net =ma. The masses cancel out leaving the acceleration of the sliding box. Find the component of the weight that pulls the box down the incline. This component is the parallel force, not the perpendicular force. If the box were freely falling, this would be true. However, in this case it is just sliding down an incline. The entire weight of the box isn t pulling it down the incline, only a component of that force. Find the force that is responsible for the box sliding down the incline. This would be the component of the weight that is parallel to the incline. Known : 40 Angle of incline Unknown :Acceleration, a =?m/s 2 Define: The component of the weight that pulls the box down the incline is the parallel force, F To find F, use opp sin θ = hyp The opposite side is our F in this case, 40 o m W=mg 40 o F F =sinθ(weight) Weight = mg F net = ma, which is equal to F F net = sin40(mg) ma= sin40(mg) Notice the mass cancels out leaving our acceleration. Output : a= sin 40(g) = x 9.8 m/s 2 = 6.3 m/s 2 = 6.3 m/s/s Substantiate: Units are correct, sig figs are correct and magnitude is reasonable. The correct answer is (B).

10 Question No. 10 of Two masses are hung over a mass-less and friction-less pulley as shown. What is the resulting acceleration of the system? Question 10 (A) 0 m/s 2 (B) 1.5 m/s 2 (C) 2.0 m/s 2 (D) 9.8 m/s 2 (E) 6.0 m/s 2 2kg 3kg The two forces pulling on each side aren t in equilibrium. There will be a net force and acceleration. Consider the two masses as forces pulling in opposite directions due to the pulley. Find the net force for this situation. C. Correct! Consider the two weights when calculating the net force. Due to the pulley, those forces oppose each other giving a net force of 29.4 N-19.6 N = 9.8 N. Here the direction of the 3 kg mass is arbitrarily made positive. The entire mass of the system must be accounted for when finding the acceleration. Use F net = ma to get an acceleration of a = F/m = 9.8/5 = 2.0 m/s 2 Although each mass if separate would fall at 9.8 m/s/s, together they do not. The 2 kg mass slows the system as a whole. When calculating the acceleration, notice that both masses will be moving. Thus, use their total mass when calculating the acceleration. Known: Mass m 1 = 2 kg Mass m 2 =3 kg Unknown: Acceleration a =? m/s 2 Define: Consider the two masses as forces pulling in opposite directions due to the pulley. Find the net force for this situation. We will arbitrarily consider the direction of the 3kg mass as positive. Find the weight of each and subtract. 2kg 3kg F net = m 2 g-m 1 g = (m 2 m 1 ) g However, when a mass is used in the next step, the entire mass of the system since the whole system will be moving N 29.4 N Output: F net = ma ( ) F m m g net 1 2 a = = m + m m + m ( m m 2 1) g ( 3-2 ) 9.8 m/s 9.8 m/s 2 a = = = = 2.0 m/s m + m Substantiate: Units are correct, sig figs are correct and magnitude is reasonable. The correct answer is (C).