Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 4, Problem 4P

Two forces, F 1 = ( 6 i ^ 4 j ^ ) N and F 2 = ( 3 i ^ + 7 j ^ ) N , act on a particle of mass 2.00 kg that is initially at rest at coordinates (−2.00 m, +4.00 m). (a) What are the components of the particle’s velocity at t = 10.0 s? (b) In what direction is the particle moving at t = 10.0 s? (c) What displacement does the particle undergo during the first 10.0 s? (d) What are the coordinates of the particle at t = 10.0 s?

(a)

Expert Solution
Check Mark
To determine

The components of the particle’s velocity.

Answer to Problem 4P

The components of the particle’s velocity is 45.0m/s , 15.0 m/s_.

Explanation of Solution

Write the expression for velocity.

v=at        (I)

Here, v is the velocity, a is the acceleration and t is the time.

Write the expression for acceleration.

a=Fnetm        (II)

Here, Fnet is the total force, m is the mass.

Write the expression for net force.

Fnet=F1+F2        (III)

Here, F1,F2 are the applied forces.

Rewrite the expression for (I) by using (II) and (III).

v=(F1+F2m)t        (IV)

Conclusion:

Substitute, (6i^4j^)N for F1, (3i^+7j^)N for F2, 2.00kg for m, 10.0s for t in equation (IV) to find v.

v=(((6i^4j^)N)+((3i^+7j^)N)(2.00kg))(10.0s)=[(4.50i^+1.50j^)m/s2](10.0s)=(45.0i^+15j^)m/s

Thus, the components of the particle’s velocity is 45.0m/s , 15.0 m/s_.

(b)

Expert Solution
Check Mark
To determine

The direction of the moving particle.

Answer to Problem 4P

The direction of the moving particle is 162° from the x-axis_.

Explanation of Solution

Write the expression for the direction of motion.

θ=180°+[tan1(vyvx)]        (V)

Here, θ is the angle, vx,vy are the x and y component of velocity.

Conclusion:

Substitute, 15.0m/s for vy, 45.0m/s for vx in equation (V) to find θ.

θ=180°+[tan1(15.0m/s45.0m/s)]=180°+(18.4°)162°from the x-axis

Thus, the direction of the moving particle is 162° from the x-axis_.

(c)

Expert Solution
Check Mark
To determine

The displacement of the particle.

Answer to Problem 4P

The displacement of the particle is (225i^+75.0j^)m_.

Explanation of Solution

Write the expression for displacement.

(Δr)=(Δx)+(Δy)        (IV)

Here, (Δr) is the displacement, (Δx) is the x-displacement, (Δy) is the y-displacement.

Write the expression for x-displacement.

(Δx)=vxit+12(axi^)t2        (V)

Here, vxi is the initial velocity along x-axis, t is the time, axi^ is the x-component of acceleration.

Write the expression for y-displacement.

(Δy)=vyit+12(ayj^)t2        (VI)

Here, vyi is the initial velocity along y-axis, t is the time, ayi^ is the y-component of acceleration.

Write the expression for the acceleration.

a=axi^+ayj^        (VII)

Rewrite the expression for the displacement by using (II), (III), (V), (VI) and (VII).

(Δr)=[vxit+12(axi^)t2]+[vyit+12(ayj^)t2]=(vxi+vyi)t+12t2(axi^+ayj^)=(vxi+vyi)t+12t2a=(vxi+vyi)t+12t2(F1+F2m)        (VIII)

Conclusion:

Substitute, (6i^4j^)N for F1, (3i^+7j^)N for F2, 2.00kg for m, 10.0s for t, 0 for vxi, 0 for vyi in equation (VIII) to find (Δr).

(Δr)=[(0+0)(10.0s)]+[12(10.0s)2([(6i^4j^)N]+[(3i^+7j^)N](2.00kg))]=[12(10.0s)2([(6i^4j^)N]+[(3i^+7j^)N](2.00kg))]=(225i^+75.0j^)m

Thus, the displacement of the particle is (225i^+75.0j^)m_.

(d)

Expert Solution
Check Mark
To determine

The coordinates of the particle.

Answer to Problem 4P

The coordinates of the particle is (227m , 79.0m)_.

Explanation of Solution

Write the expression for position of the particle.

rf=ri+(Δr)        (IX)

Here, rf is the final position, ri is the initial position.

Conclusion:

Substitute, (2.00i^+4.00j^)m for ri, (225i^+75j^)m for (Δr) in equation (IX) to find rf.

rf=((2.00i^+4.00j^)m)+((225i^+75.0j^)m)=(227i^+79.0j^)m

Thus, the coordinates of the particle is (227m , 79.0m)_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A ball of mass 0.3 kg, initially at rest, is kicked directly toward a fence from a point 20 m away, as shown below. The velocity of the ball as it leaves the kicker’s foot is 17 m/s at angle of 47 ◦ above the horizontal. The top of the fence is 6 m high. The ball hits nothing while in flight and air resistance is negligible. The acceleration due to gravity is 9.8 m/s2 (meters/seconds/squared). 1) How far above the top of fence will the ball pass? Consider the diameter of the ball to be negligible. Answer in units of m. 2) What is the vertical component of the velocity when the ball reaches the plane of the fence? Answer in units of m/s.
In the figure, a body with a mass of 2 kg moves under the influence of two constant forces F1 = 5N and F2 = 4N in the xy plane. At time t = 0, the object is at point 0 and its speed is V = 2i + j (m / s). What is the acceleration of the particle and its position after 2 seconds in terms of the unit vector?
The position vector of a particle of mass 0.05 kg is r (t) = 3 cos (4. t)i + 3 sin (4.t)j +10 t k What is the vector expression for the velocity of the particle at time t in m/s? v (t) = ? What is the vector expression for the acceleration of the particle at time t in m/s^2? a(t) = ? Deduce the magnitude of the force acting on the particle |F| = What is the direction of the acceleration compared with the velocity of the particle?

Chapter 4 Solutions

Principles of Physics: A Calculus-Based Text

Ch. 4 - Prob. 4OQCh. 4 - Prob. 5OQCh. 4 - Prob. 6OQCh. 4 - Prob. 1CQCh. 4 - If a car is traveling due westward with a constant...Ch. 4 - A person holds a ball in her hand. (a) Identify...Ch. 4 - Prob. 4CQCh. 4 - If you hold a horizontal metal bar several...Ch. 4 - Prob. 6CQCh. 4 - Prob. 7CQCh. 4 - Prob. 8CQCh. 4 - Balancing carefully, three boys inch out onto a...Ch. 4 - Prob. 10CQCh. 4 - Prob. 11CQCh. 4 - Prob. 12CQCh. 4 - Prob. 13CQCh. 4 - Give reasons for the answers to each of the...Ch. 4 - Prob. 15CQCh. 4 - In Figure CQ4.16, the light, taut, unstretchable...Ch. 4 - Prob. 17CQCh. 4 - Prob. 18CQCh. 4 - Prob. 19CQCh. 4 - A force F applied to an object of mass m1 produces...Ch. 4 - (a) A car with a mass of 850 kg is moving to the...Ch. 4 - A toy rocket engine is securely fastened to a...Ch. 4 - Two forces, F1=(6i4j)N and F2=(3i+7j)N, act on a...Ch. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Two forces F1 and F2 act on a 5.00-kg object....Ch. 4 - A 3.00-kg object is moving in a plane, with its x...Ch. 4 - A woman weighs 120 lb. Determine (a) her weight in...Ch. 4 - Prob. 10PCh. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - You stand on the seat of a chair and then hop off....Ch. 4 - Prob. 17PCh. 4 - A block slides down a frictionless plane having an...Ch. 4 - Prob. 19PCh. 4 - A setup similar to the one shown in Figure P4.20...Ch. 4 - Prob. 21PCh. 4 - The systems shown in Figure P4.22 are in...Ch. 4 - A bag of cement weighing 325 N hangs in...Ch. 4 - Prob. 24PCh. 4 - In Example 4.6, we investigated the apparent...Ch. 4 - Figure P4.26 shows loads hanging from the ceiling...Ch. 4 - Prob. 27PCh. 4 - An object of mass m1 = 5.00 kg placed on a...Ch. 4 - An object of mass m = 1.00 kg is observed to have...Ch. 4 - Two objects are connected by a light string that...Ch. 4 - Prob. 31PCh. 4 - A car is stuck in the mud. A tow truck pulls on...Ch. 4 - Two blocks, each of mass m = 3.50 kg, are hung...Ch. 4 - Two blocks, each of mass m, are hung from the...Ch. 4 - In Figure P4.35, the man and the platform together...Ch. 4 - Two objects with masses of 3.00 kg and 5.00 kg are...Ch. 4 - A frictionless plane is 10.0 m long and inclined...Ch. 4 - Prob. 39PCh. 4 - An object of mass m1 hangs from a string that...Ch. 4 - A young woman buys an inexpensive used car for...Ch. 4 - A 1 000-kg car is pulling a 300-kg trailer....Ch. 4 - An object of mass M is held in place by an applied...Ch. 4 - Prob. 44PCh. 4 - An inventive child named Nick wants to reach an...Ch. 4 - In the situation described in Problem 45 and...Ch. 4 - Two blocks of mass 3.50 kg and 8.00 kg are...Ch. 4 - Prob. 48PCh. 4 - In Example 4.5, we pushed on two blocks on a...Ch. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52PCh. 4 - Review. A block of mass m = 2.00 kg is released...Ch. 4 - A student is asked to measure the acceleration of...Ch. 4 - Prob. 55PCh. 4 - Prob. 56PCh. 4 - A car accelerates down a hill (Fig. P4.57), going...Ch. 4 - Prob. 58PCh. 4 - In Figure P4.53, the incline has mass M and is...
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Electric Fields: Crash Course Physics #26; Author: CrashCourse;https://www.youtube.com/watch?v=mdulzEfQXDE;License: Standard YouTube License, CC-BY