Newton's 2nd Law stipulates that "the sum of all forces acting on an object are equal to the mass times the acceleration of that object." Given the force diagram below, calculate the acceleration of the object (both magnitude and direction). The object is the smiley face, which has a mass of 18.2 kg (it's a very heavy smiley face). FPush = 42.0 N Fg = Weight
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- Jason is pulling a box across the room. He is pulling with a force of 24 newtons and his arm is making a 63 angle with the horizontal, if the box weighs 23 newtons what is the net force on the box in the vertical direction? Treat up as the positive direction, and down as the negative direction. Answer valueA constant net force causes an acceleration of 20 m/s². What will be the acceleration of the object if the force is halved? Enter to 2 significant figures m/s² a = A constant net force causes an acceleration of 20 m/s². What will be the acceleration of the object if the mass is doubled? Enter to 2 significant figures m/s² a =A 100. N force is applied to a 21.2 kg crate to slide it along a rough horizontal surface. The diagram shows the forces acted on the crate in this situation. What was the magnitude of the normal force acting on the crate? (a) 208 N (b) 258 N (c) 121 N (d) 69.6 N (e) 158 N
- An applied force of 89.4 N is used to accelerate an object across a frictional surface. The object has aweight of 61.2 N and encounters a frictional force of 76.7 N.(a) Draw a sketch of the object and show all of the forces acting on the object.(b) What is the net force acting on the object in the horizontal direction?(c) What is the net force acting on the object in the vertical direction?(d) Using Newton's 2nd Law, find the object's acceleration.Block A has a mass of 10.0 kg, block B has a mass of 11.0 kg, block C has a mass of 12.0 kg. If there is an applied force of 18.0 N, find the magnitude of the tension in the rope between blocks A and B Ignore friction. (Input your answer with 1 decimal place) A B TAs shown in the figure, a woman is straining to lift a large crate, but without success because it is too heavy. We denote the forces on the crate as follows: P is the magnitude of the upward force being exerted on the crate by the person, C is the magnitude of the vertical contact force on the crate by the floor, and W is the weight of the crate. How are the magnitudes of these forces related while the person is trying unsuccessfully to lift the crate? OA P=C OB.P+C>W OCP+CThree mass are arranged as shown. Mass A is 800 kg, Mass B is 400 kg, and Mass C is 600 kg. Draw a free body diagram for each mass. Find the net, vector force on each mass. A 8 m 90° В Mass A Mass B Mass C 12 mThe drawing shows two books at rest on top of a desk. The history book is on top of the physics book, as shown. Labels for the magnitudes of relevant forces are listed below in alphabetic order. 1. FDG: the force the desk (D) exerts on the ground (G). 2. Fóp: the force the desk (D) exerts on the physics book (P). 3. FGD: the force the ground (G) exerts on the desk (D). 4. FHP: the force the history book (H) exerts on the physics book (P). 5. FpD: the force the physics book (P) exerts on the desk (D). 6. FPH: the force the physics book (P) exerts on the history book (H). 7. Wp: the weight of the desk (D). 8. WH: the weight of the history book (H). 9. Wp: the weight of the physics book (P). Begin by creating a free-body diagram for each of the objects, and then answer the questions below. ✓ 50% Part (a) All of the following relationships are correct. Which are correct due to Newton's Third Law? FDG = FGD, FPD = FDP, FHP = FPH ✓ Correct! All of the following relationships are correct.…Two people play tug of war. The 100-kg person on the left pulls with 1100 N, and the 75-kg person on the right pulls with 870 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons? Add your answer