A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. a. Express the two forces in in unit-vector notation. Ans. F = (20.81 i + 18.74 j)N, F, = (-4.45 i + 12.22 j)N b. Find the total force exerted on the cart. Ans. EF = (16.36 i + 30.96 j)N or 35.02 N Find the angle at which the total force is applied. Ans. 62.15° NEERIN

A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. a. Express the two forces in in unit-vector notation. Ans. F = (20.81 i + 18.74 j)N, F, = (-4.45 i + 12.22 j)N b. Find the total force exerted on the cart. Ans. EF = (16.36 i + 30.96 j)N or 35.02 N Find the angle at which the total force is applied. Ans. 62.15° NEERIN

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Introduction And Vectors

Section: Chapter Questions

Problem 4OQ: What is the y component of the vector (3i8k) m/s? (a) 3 m/s (b) 8 m/s (c) 0 (d) 8 m/s (e) none of...

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A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. a. Express the two forces in in unit-vector notation. Ans. F = (20.81 i+ 18.74 j)N, F, =(-4.45 i+ 12.22 j)N b. Find the total force exerted on the cart. Ans. EF = (16.36 i + 30.96 j)N or 35.02 N c. Find the angle at which the total force is applied. Ans. 62.15º d. What is the acceleration of the cart? Ans. å = (2.05 i + 3.87 j)" or 4.38 EERIN
A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. a. Express the two forces in in unit-vector notation. Ans. F = (20.81 i + 18.74 j)N, F, = (-4.45 i + 12.22 j)N b. Find the total force exerted on the cart. Ans. EF = (16.36 i+ 30.96 j)N or 35.02 N c. Find the angle at which the total force is applied. Ans. 62.15°
1. A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 m/s. a. Express the two forces in in unit-vector notation. Ans. F = (20.81 i+ 18.74 j)N, F, = (-4.45 i + 12.22 j)N b. Find the total force exerted on the cart. Ans. E F = (16.36 i + 30.96 j)N or 35.02 N c. Find the angle at which the total force is applied. Ans. 62.15° EERIN
A cart of mass 8.0kg was moved by applying two constant forces. Force 1 is 28.0N at 42.0 degrees, and Force 2 is 13.0N at 110 degrees. Initially, the cart has a velocity of (3.50i + 2.20j)m/s. a. Express the two forces in unit-vector notation.b. Find the total force exerted on the cart.c. Find the angle at which the total force is applied.d. What is the acceleration of the cart?e. What is the cart's velocity after 5.0s?f. What is the position of the cart after 5.0s?g. What is the angle of the position after 5.0s?h. What is the total work done by the two applied forces?
A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0 N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. d. What is the acceleration of the cart? Ans. ä = (2.05 i + 3.87 j) or 4.38 e. What is the cart's velocity after 5.00 s? Ans. i = (13.75 i + 21.55 j)" or 25.56" f. What is the position of the cart after 5.00 s? † = 43.125i + 59.375j or73.38 m
A cart of mass 8.00 kg was moved by applying two constant forces. Force 1 is 28.0N at 42.0°, and Force 2 is 13.0 N at 110°. Initially, the cart has a velocity of (3.50 i +2.20 j) m/s. EERIN d. What is the acceleration of the cart? Ans. å = (2.05 i + 3.87 j) e. What is the cart's velocity after 5.00 s? or 4.38- Ans. 3 = (13.75 i + 21.55 j)" or 25.56 m m
A boulder is being pushed by two different people on a flat terrain. The net force on the boulder can be expressed in the Cartesian coordinate system as vector f = (-4i + 5.5j)N. Where I and j denote directions to the east and north. A. Find the magnitude in N B. Find the direction (in degrees counterclockwise from the east axis) of the push.
Mr. One is pushing a cart with a force of E1 = 15.00 N due north. Mr. Two is pushing the same cart with a force of F2 = 25.00 N due northeast. Mr. Three is pushing the cart a force F3 so that the cart is in equilibrium. Find force F3 so that F1 + F2+F3 = 0. First, we calculate the vector sum F + F, by component method. The x-component of F1 is 0 and the y-component is 15 The x-component of F2 is 0 and the y-component is 25 Adding the components, we can express the resultant vector in unit vector component form: F+F2 = Since we want to find F1 +F2+ F3= 0, it follows that F3= - (F+F).TI Thus,
Mr. One is pushing a cart with a force of F1 - 15.00 N due north. Mr. Two is pushing the same cart with a force of F2 - 25.00 N due northeast. Mr. Three is pushing the cart a force F3 so that the cart is in equilibrium. Find force F3 so that F +Fz+F3=0. First, we calculate the vector sum F + F2 by component method. The x-component of F, is and the y-component is The x-component of Fz is and the y-component is Adding the components, we can express the resultant vector in unit vector component form: Since we want to find F +F;+F3-0, t follows that F- - (Fj +F). Thus, The magnitude of Fg is F3- N.
6. A2 kg ball was thrown in the air from an angle of 30° moving with a velocity v=20 m/s. Assume the initial starting point is on the ground. a. Illustrate a diagram of the scenario. b. Compute for the components of velocity. nc. Compute the half time of flight. imiz ez d. Compute for the maximum height. e. Find the range of the trajectory.
2.6.2 Force Resultant in 3-D F₂ = 2 KN 60 60° Express each force in Cartesian vector form. 45 F₁ = 5 KN O F₁ = (2.50 i +3.54 j +2.50 k) kN, F₂ = −2 j kN Ο O F₁ = (2.17i+3.75 j +4.33 k) kN, F₂ = -2 j kN F1 O F₁ = (4.33 i +3.75j +2.17 k) kN, F₂ = -2 j kN O F₁ = (4.33 i +3.54j+4.33 k) kN, F₂ = -2 j kN
John fires a projectile in a Uwindsor lab and notes that the 20.0g projectile travels 785 m/s. It then strikes a ball of clay and penetrates 24.5 cm into it. Based on this information, what 48. would John calculate as the average force the projectile exerts on the clay? А. 2.07 x 105 N В. 1.97 x 104 N С. 3.93 х 104 N D. 2.52 x 104 N Е. 6.26 х 103 N