Skip to main content
Ch 05: Applying Newton's Laws
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610당신이 사용하는 게 아니라요?교과서 변경
5장, 문제 29b

A 45.045.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313313 N. Then you must reduce your push to 208208 N to keep it moving at a steady 25.025.0 cm/s. What push must you exert to give it an acceleration of 1.101.10 m/s2?

검증된 단계별 안내
1
Step 1: Identify the forces acting on the crate. The forces include the applied force (F), the force of kinetic friction (F_k), and the net force (F_net). The force of kinetic friction is given as 208 N when the crate is moving at a constant velocity.
Step 2: Use Newton's second law of motion, which states that F_net = m * a, where m is the mass of the crate (45.0 kg) and a is the acceleration (1.10 m/s²). The net force is the difference between the applied force (F) and the force of kinetic friction (F_k).
Step 3: Write the equation for the net force: F_net = F - F_k. Substitute F_net = m * a into this equation to get F - F_k = m * a.
Step 4: Rearrange the equation to solve for the applied force (F): F = F_k + m * a. Substitute the known values: F_k = 208 N, m = 45.0 kg, and a = 1.10 m/s².
Step 5: Perform the calculation to find the applied force (F). This will give you the force required to accelerate the crate at 1.10 m/s² while overcoming the force of kinetic friction.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
3m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship is expressed by the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. Understanding this law is crucial for determining the force required to accelerate the crate.
추천 영상:
06:54
Intro to Forces & Newton's Second Law

Friction

Friction is the force that opposes the relative motion of two surfaces in contact. In this scenario, static friction prevents the crate from moving until a certain threshold force is applied, while kinetic friction acts on the crate once it is in motion. The coefficients of static and kinetic friction are essential for calculating the forces involved in moving the crate.
추천 영상:
08:11
Static Friction & Equilibrium

Net Force

The net force is the vector sum of all the forces acting on an object. In the context of the crate, the net force is the difference between the applied force and the frictional force opposing the motion. To achieve a specific acceleration, the applied force must overcome both static and kinetic friction, resulting in the necessary net force to produce the desired acceleration.
추천 영상:
06:28
Finding Net Forces in 2D Gravitation
관련 실천
교과서 질문

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level bed of the truck are 0.3550.355 and 0.6500.650, respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to 30.030.0 m/s without causing the box to slide? Draw a free-body diagram of the toolbox.

2662
views
2
rank
교과서 질문

A box of bananas weighing 40.0 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.400.40, and the coefficient of kinetic friction is 0.200.20. What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

1710
views
교과서 질문

A box of bananas weighing 40.040.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.400.40, and the coefficient of kinetic friction is 0.200.20. If the monkey applies a horizontal force of 18.018.0 N, what is the magnitude of the friction force and what is the box's acceleration?

1715
views
교과서 질문

A 45.045.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313313 N. Then you must reduce your push to 208208 N to keep it moving at a steady 25.025.0 cm/s. Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is 1.621.62 m/s2.

(i) What magnitude push would cause it to move?

(ii) What would its acceleration be if you maintained the push in part (b)? Note: Part (b) asked what push you must exert to give it an acceleration of 1.101.10 m/s2.

1194
views
교과서 질문

A 25.025.0-kg box of textbooks rests on a loading ramp that makes an angle αα with the horizontal. The coefficient of kinetic friction is 0.250.25, and the coefficient of static friction is 0.350.35. At this angle, find the acceleration once the box has begun to move.

2513
views
교과서 질문

A 45.045.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313313 N. Then you must reduce your push to 208208 N to keep it moving at a steady 25.025.0 cm/s. What are the coefficients of static and kinetic friction between the crate and the floor?

3347
views