# Newton's Laws Review

Unit 2

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### Part F: Problem-Solving

For the following problems, draw free-body diagrams and solve for the requested unknown. Use g = 9.8 m/s/s.

Free-Body Diagrams | Finding Acceleration || Finding Individual Forces | Kinematic Equations and Problem-Solving

36. A 0.25-kg ball is thrown upwards with an initial velocity of 12 m/s at an angle of 30 degrees. Determine the acceleration of the ball when it has reached the peak of its trajectory. Assume negligible air resistance. PSYW

 Answer: 9.8 m/s/s, down There is only one force upon the ball - the force of gravity. (Air resistance is neglible; the ball is not on a surface, so there is no friction or normal force; the applied force which projects it into motion does not act upon the ball during its trajectory; there are no springs, strings, wires, or cables attached so there is neither a tension force nor a spring force.) The force of gravity acts downward with a magnitude of m•g = (0.25 kg) •(9.8 m/s/s) = 2.45 N. The net force is 2.45 N; when divided by mass, the acceleration can be found.  a = Fnet / m = (2.45 N, down) / (0.25 kg) = 9.8 m/s/s, down

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37. A 72-kg skydiver is falling from 10000 feet. At an instant during the fall, the skydiver encounters an air resistance force of 540 Newtons. Determine the acceleration of the skydiver at this instant. PSYW

 Answer: 2.30 m/s/s, down There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. The sum of the vertical forces is ·Fy = 540 N, up + 706 N, down = 166 N, down  The acceleration of the skydiver can be computed using the equation ·Fy = m•ay. ay = (166 N, down) / (72 kg) = 2.30 m/s/s, down

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38. A 72-kg skydiver is falling from 10 000 feet. After reaching terminal velocity, the skydiver opens his parachute. Shortly thereafter, there is an an instant in time in which the skydiver encounters an air resistance force of 1180 Newtons. Determine the acceleration of the skydiver at this instant. PSYW

 Answer: 6.59 m/s/s, up There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. The sum of the vertical forces is ·Fy = 1180 N, up + 706 N, down = 474 N, up  The acceleration of the skydiver can be computed using the equation ·Fy = m•ay. ay = (474 N, up) / (72 kg) = 6.59 m/s/s, up

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39. A 5.2-N force is applied to a 1.05-kg object to accelerate it rightwards across a friction-free surface. Determine the acceleration of the object. (Neglect air resistance.) PSYW

 Answer: 4.95 m/s/s, right Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 5.2 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.  a = Fnet / m = (5.2 N, right) / (1.05kg) = 4.95 m/s/s, right

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40. A 5.2-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine the acceleration of the object. (Neglect air resistance.) PSYW

 Answer: 1.82 m/s/s, right Upon neglecting air resistance, there are four forces acting upon the object. The up and down forces balance each other. The acceleration is rightward since the rightward applied force is greater than the leftward friction force. The horizontal forces can be summed as vectors in order to determine the net force. Fnet = ·Fx = 5.2 N, right - 3.29 N, left = 1.91 N, right The acceleration of the object can be computed using Newton's second law.  ax = ·Fx / m = (1.91 N, down) / (1.05kg) = 1.82 m/s/s, right

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41. A 1250-kg small aircraft decelerates along the runway from 36.6 m/s to 6.8 m/s in 5.1 seconds. Determine the average resistive force acting upon the plane. (Assume that its engine/propeller makes no contributes to its forward motion). PSYW

 Answer: 7304 N This problem is similar to the two previous problems in many respects: the free-body diagram is identical or similar and the acceleration is not given but determinable from the kinematic information. vi = 36.6 m/s, vf = 6.8 m/s, and t = 5.1 s  The acceleration of the object is the velocity change per time: a = Delta v / t = (6.8 m/s - 36.6 m/s) / (5.1 s) = -5.84 m/s/s or 5.84 m/s/s, left. This acceleration can be used to determine the net force: Fnet = m•a = (1250 kg) • (5.84 m/s/s, left) = 7304 N, left The friction forces (surface and air) provide this net force and are equal in magnitude to this net force.

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42. A tow truck exerts a 18300-N force upon a 1200-kg car to drag it out of a mud puddle onto the shoulder of a road. A 17900 N force opposes the car's motion. The plane of motion of the car is horizontal. Determine the time required to drag the car a distance of 6.9 meters from its rest position. PSYW

 Answer: 6.43 s Upon neglecting air resistance, there are four forces acting upon the object. The up and down forces balance each other. The acceleration is rightward (or in the direction of the applied force) since the rightward applied force is greater than the leftward friction force. The horizontal forces can be summed as vectors in order to determine the net force. Fnet = ·Fx = 18300 N, right - 17900 N, left = 400 N, right The acceleration of the object can be computed using Newton's second law.  ax = ·Fx / m = (400 N, down) / (1200 kg) = 0.333 m/s/s, right This acceleration value can be combined with other known kinematic information (vi = 0 m/s, d = 6.9 m) to determine the time required to drag the car a distance of 6.9 m. The following kinematic equation is used; substitution and algebra steps are shown. d = vi • t + 0.5 •a • t2 d = vi • t + 0.5 •a • t2 6.9 m = 0.5 • (0.333 m/s/s) • t2 6.9 m / (0.5 • 0.333 m/s/s ) = t2 41.4 = t2 6.43 s = t

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43. A 4.44-kg bucket suspended by a rope is accelerated upwards from an initial rest position. If the tension in the rope is a constant value of 83.1 Newtons, then determine the acceleration (in m/s/s) of the bucket. PSYW

 Answer: 14.2 m/s There are two forces acting upon the bucket - the force of gravity (up) and the tension force (down). The magnitude of the force of gravity is found from m•g; its value is 43.5 N. These two forces can be summed as vectors to determine the net force. Fnet = ·Fy = 83.1 N, up + 43.5 N, down = 39.6 N, up The acceleration can be calculated using Newton's second law of motion. a = / m = (39.6 N, up) / (4.44 kg) = 8.92 m/s/s, up

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44. A shopper in a supermarket pushes a loaded cart with a horizontal force of 16.5 Newtons. If the cart has a mass of 33.8 kg, how far (in meters) will it move in 1.31 seconds, starting from rest? (Neglect resistive forces.)

 Answer: 0.419 m Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 22.6 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.  a = Fnet / m = (16.5 N, right) / (33.8 kg) = 0.488 m/s/s, right The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 1.31 s) to calculate the final speed of the cart. The kinematic equation, substitution and algebra steps are shown. d = vi • t + 0.5 •a • t2 d = vi • t + + 0.5 • (0.488 m/s/s)•(1.31 s) 2 d = 0.419 m

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Free-Body Diagrams | Finding Acceleration || Finding Individual Forces | Kinematic Equations and Problem-Solving