Concept explainers
Compare the net force on a heavy trunk when it’s (a) at rest on the floor; (b) being slid across the floor at constant speed; (c) being pulled upward in an elevator whose cable tension equals the combined weight of the elevator and trunk; and (d) sliding down a frictionless ramp.
(a)
Answer to Problem 1FTD
Explanation of Solution
According to Newton’s first law of motion a body will remain in rest until it is acted upon by external unbalanced force. Therefore the net force acting on a body at rest is zero.
The forces acting on a body are weight of the body which is down ward and normal reaction which is upward. For a body at rest on the floor therefore the normal reaction will be equal to that of weight of the body. Thus net force will be zero.
Conclusion:
Thus, the net force on a heavy trunk when it is at rest on the floor is zero.
(b)
Answer to Problem 1FTD
Explanation of Solution
According to Newton’s first law of motion a body will remain in uniform motion until it is acted upon by external unbalanced force. Therefore the net force acting on a body in uniform motion is zero.
The forces acting on the trunk are the weight of the trunk which is acting downward and normal reaction which is acting upward, driving force along the direction of motion, air resistance opposite to the direction of force and friction opposite to the direction of motion. The normal reaction balances weight of the trunk and driving force balances the sum of air resistance and friction. Thus net force acting on a constant speed car is zero.
Conclusion:
Thus, the net force on a heavy trunk when it is being slid across the floor at constant speed is zero.
(c)
Answer to Problem 1FTD
Explanation of Solution
According to Newton’s first law of motion a body will remain in uniform motion until it is acted upon by external unbalanced force.
The forces acting on the elevator and trunk is the weight of the body acting downward and tension of the cable acting upward. Since both are same net force acting on the system is zero. Thus total acceleration is zero. Therefore net force on elevator and trunk is zero.
Conclusion:
Thus, the net force on a heavy trunk when it is being pulled upward in an elevator whose cable tension equals the combined weight of the elevator and trunk is zero
(d)
Answer to Problem 1FTD
Explanation of Solution
According to Newton’s first law of motion a body will move with acceleration only if there is any net force acting on it.
The trunk sliding down is similar to free fall. The only force is the weight of the body which is downward. Therefore net force on the trunk is downward.
Conclusion:
Thus, the net force on a heavy trunk when it is sliding down a frictionless ramp is downward.
Want to see more full solutions like this?
Chapter 5 Solutions
Essential University Physics: Volume 1 (3rd Edition)
Additional Science Textbook Solutions
Microbiology with Diseases by Body System (5th Edition)
Anatomy & Physiology (6th Edition)
Campbell Biology (11th Edition)
Introductory Chemistry (6th Edition)
Human Anatomy & Physiology (2nd Edition)
Cosmic Perspective Fundamentals
- How can i solve this if n1 (refractive index of gas) and n2 (refractive index of plastic) is not known. And the brewsters angle isn't knownarrow_forward2. Consider the situation described in problem 1 where light emerges horizontally from ground level. Take k = 0.0020 m' and no = 1.0001 and find at which horizontal distance, x, the ray reaches a height of y = 1.5 m.arrow_forward2-3. Consider the situation of the reflection of a pulse at the interface of two string described in the previous problem. In addition to the net disturbances being equal at the junction, the slope of the net disturbances must also be equal at the junction at all times. Given that p1 = 4.0 g/m, H2 = 9.0 g/m and Aj = 0.50 cm find 2. A, (Answer: -0.10 cm) and 3. Ay. (Answer: 0.40 cm)please I need to show all work step by step problems 2 and 3arrow_forward
- look at answer show all work step by steparrow_forwardLook at the answer and please show all work step by steparrow_forward3. As a woman, who's eyes are h = 1.5 m above the ground, looks down the road sees a tree with height H = 9.0 m. Below the tree is what appears to be a reflection of the tree. The observation of this apparent reflection gives the illusion of water on the roadway. This effect is commonly called a mirage. Use the results of questions 1 and 2 and the principle of ray reversibility to analyze the diagram below. Assume that light leaving the top of the tree bends toward the horizontal until it just grazes ground level. After that, the ray bends upward eventually reaching the woman's eyes. The woman interprets this incoming light as if it came from an image of the tree. Determine the size, H', of the image. (Answer 8.8 m) please show all work step by steparrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-HillUniversity Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice University