MCAT Physics Problem Solving Drill 07: Force and Motion


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1 MCAT Physics Problem Solving Drill 07: orce and Motion Question No. 1 of 10 Question 1. A 10 kg sled is pushed at a constant velocity of 4 m/s along a snowy surface. A frictional force of 10 N acts on it. What is the net force acting on the sled? Question #01 (A) 196 N (B) 108 N (C) 98 N (D) 0 N Consider all forces acting on the sled to obtain the net force. Be sure to consider their direction too. B. Incorrect. 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. D. 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.
2 Question No. 2 of 10 Question 2. A 40 kg crate is pushed with a force of 100 N along a horizontal surface. The crate moves at a constant velocity. What is the coefficient of friction between these moving surfaces? Question #02 (A) 0.4 (B) 0.5 (C) 0.26 (D) 0 Don t confuse the mass and weight. They aren t the same quantity. B. Incorrect. The coefficient of friction is the frictional force divided by the normal force. The frictional force must be equal in size to the applied force on the object. C. Correct! The coefficient of friction is the frictional force divided by the normal force. 100N divided by 392 N equals.26. There must be some friction in this case. Otherwise the crate would be accelerating since there is a force acting on it. The coefficient of friction is the frictional force divided by the normal force. f μ k = N In this case, the normal force is equal to the objects weight. The frictional force is equal to the applied force on the object. Since it is moving at a constant velocity, all these forces give a net force of zero on the object. 100N μ k = =.26 2 (40kg)(9.8m/s ) Be sure to use the weight of the object, not the mass. Also, notice that the coefficient of friction is a unitless quantity.
3 Question No. 3 of 10 Question 3. A box slides down a frictionless inclined plane. It makes an angle of 30 with the horizontal. What is the acceleration of the box as it slides down the incline plane. Question #03 (A) 0 m/s 2 (B) 5 m/s 2 (C) 7.5 m/s 2 (D) 9.8 m/s 2 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! ind 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. net = ma. The masses cancel out leaving the acceleration of the sliding box. ind 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. ind 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. sin θ=opp/hyp The opposite side is our in this case. =(sin θ) (weight) =(sin30 ) (mg) net =(sin30 ) (mg) Since this is the net force, we can also set this equal to ma. ma = (sin 30 ) mg Notice the mass cancels out leaving our acceleration. a = (sin 30 ) mg = m/s 2 = 5 m/s 2 m 30 o W=mg 30 o
4 Question No. 4 of 10 Question 4. Two masses are hung over a massless and frictionless pulley as shown. What is the resulting acceleration of the system? Question #04 (A) 0 m/s 2 (B) 2 m/s 2 (C) 3 m/s 2 (D) 10 m/s 2 2kg 3kg The two forces pulling on each side aren t in equilibrium. There will be a net force and acceleration. B. Correct! Consider the two weights when calculating the net force. Due to the pulley, those forces oppose each other giving a net force of (30 20) N = 10 N. Here the direction of the 3kg mass is arbitrarily made positive. The entire mass of the system must be accounted for when finding the acceleration. Use net = (3 kg + 2 kg) a to get an acceleration of 2 m/s 2. The net force must be divided by the sum of the masses to get the acceleration; don t just divide the net force by the 3 kg mass. 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. Consider the two masses as forces pulling in opposite directions due to the pulley. ind the net force for this situation. We will arbitrarily consider the direction of the 3kg mass as positive. net = 30 N 20 N = 10 N However, when a mass is used in the next step, be sure to use the entire mass of the system since the whole system will be moving. net = ma 10 N = (5 kg) a a= 2 m/s 2 2kg 3kg 20 N 30 N
5 Question No. 5 of 10 Question 5. Consider a 100 N weight hanging from two cables attached to walls as shown. Calculate the tension/force in the second cable, T2. Question #05 (A) 115 N (B) 100 N (C) 145 N (D) 200 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 100 N to balance the hanging weight. ind the hypotenuse, which is the force on T2. cos 30 = 100 N / T2; solving for T2 gives 115 N. B. Incorrect. Consider the tension in T2 as horizontal and vertical components. Only the vertical component of T2 would equal 100 N. 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 100 N since that must equal the hanging weight. 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. cos adj θ = cos hyp 100 N θ = T2 100 N T 2 = = 115 N o cos o T2 T1 100N
6 Question No. 6 of 10 Question 6. A 10 kg package is pulled straight upward with 125 N of force. What is the magnitude of the acceleration of the package? Question #06 (A) 0 m/s 2 (B) 2.5 m/s 2 (C) 12.5 m/s 2 (D) 22.5 m/s 2 The net force on the object is upward, so the acceleration will be nonzero. B. Correct! The downward force on the object is W = (10 m/s 2 ) 10 kg = 100 N. The net force is 25 N upward, so the acceleration is 2.5 m/s 2. See the solution below for a more detailed answer. This is the acceleration that the object would undergo in the absence of the force of gravity. Although the total force applied on the object is 225 N, force is a vector, and must be added as such. irst, calculate the net force. Next, divide the net force by the mass to find the acceleration. net =mg=( ) N=25 N m 25 N 10 kg net 2 a= = = 2.5 m/s
7 Question No. 7 of 10 Question 7. A 2.0 kg mass is being pulled across the floor. If the coefficient of friction is 0.15, and the mass is accelerating at 1.0 m/s 2, what is the magnitude of the force pulling the object? Question #07 (A) 1.0 N (B) 2.0 N (C) 3.0 N (D) 5.0 N The force of friction is directed opposite to the force pulling the object. B. Incorrect. The force pulling the object is not just the mass multiplied by the acceleration. The question is not asking for the net force. The force required to pull the object is not just that required to counteract the force of friction; in that case, the object would not accelerate at all. D. Correct! irst, compute the net force on the object. The force pulling the object is the net force plus the force of friction directed in the opposite direction. irst, write the net force as the force applied minus the force of friction, since friction is directed opposite to the force applied. Next, substitute ma as the value of the net force and μmg as the value of the force applied. Then add these values together to obtain the final answer. net=a f ma=a μmg =ma+μmg A = N =2+3 N ( ) ( ) =5 N
8 Question No. 8 of 10 Question 8. A car with a mass of 1000 kg is driven up a road with a 30.0 slope. Assuming the coefficient of friction is 0.20, what force must the car exert to move with constant velocity? Question #08 (A) 0 N (B) 1700 N (C) 5000 N (D) 6700 N Although the net acceleration and net force is zero, the car must exert a force to counteract the component of gravity parallel to the road and the force of friction. B. Incorrect. Although this is the force required to counteract friction, the force of gravity s component parallel to the road must also be accounted for. This is the force that the car would exert in the absence of friction. D. Correct! irst, compute the net force on the object, which is zero due to no acceleration. The net force is the force applied, A, minus the force of friction, f, minus the component of the force of gravity parallel to the road, m x g x cos θ. Solve for A, as shown in the detailed solution below. irst, write the net force as the force applied minus the force of friction minus the component of gravity parallel to the road. The net force is zero, the force of friction is f = μ N = μmgcosθ, and the component of the force of gravity parallel to the road is mgsinq. Plug these values in, and then solve for the force applied, A. The work is shown below. net=a mgcosθ f 0=A μmgcosθmgsinθ =μmgcosθ+mgsinθ A = μcosθ+sinθ mg ( ) 2 ( ) ( ) ( ) = kg 10 m/s ( ) = = 6700 N
9 Question No. 9 of 10 Question 9. A block slides down an inclined plane at constant velocity. What is the coefficient of kinetic friction between the block and the plane? 2 1 Question #09 3 (A) 1/2 (B) 3/3 (C) 3/2 (D) 3 We are looking for the tangent of the angle of inclination, not the sine of the angle of inclination of the plane. B. Correct! The coefficient of friction is given by the tangent of the angle of inclination of the plane. We are looking for the tangent of the angle of inclination, not the cosine of the angle of inclination of the plane. This is the reciprocal of the answer we are looking for. Make sure you have identified the correct angle. irst, write the net force as the component of the force of gravity parallel to the road minus the force of friction. Since the velocity is constant, there is no acceleration; therefore, the net force is zero. The force of friction is f = μ N = μmgcosθ, and the component of the force of gravity parallel to the road is m g sinθ. Plug these values in, and then solve for μ., as shown below. net=mgsinθμmgcosθ 0=mgsinθμmgcosθ μmgcosθ=mgsinθ μcosθ=sinθ sinθ μ= cosθ =tanθ opp = adjacent 1 = 3 3 = 3
10 Question No. 10 of 10 Question 10. A 70 kg person is standing on a scale in an elevator, which is accelerating downward at 2 m/s 2. What value does the scale read? Question #10 (A) 14 kg (B) 56 kg (C) 70 kg (D) 84 kg Although the scale reading is reduced, this is not the correct value. B. Correct! The net force on the person is the normal force minus the force of gravity. The net force is equal to the acceleration (2 m/s²) multiplied by the mass of the person. Solve for the normal force, and divide it by the acceleration of gravity to get the mass reading on the scale. Due to the acceleration on the person, the scale reading will be different. Use Newton s Laws to solve for the normal force on the person. This would be the scale reading if the elevator were accelerating upward at 2 m/s 2. irst, write the net force on the person as the force of gravity minus the normal force The net force is equal to mass, m, multiplied by acceleration, a = 2 m/s 2. Solve for the normal force, which is equal in magnitude to the force the person exerts on the scale. The normal force divided by g gives the scale reading. The work is shown below. net=mg N=ma N=m(ga) m'g=m(ga) ga m'=m g 102 = ( 70 kg) 10 = 56 kg
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