*expand_more*

*expand_more*

*format_list_bulleted*

Figure 5-19 gives the free-body diagram for four situations in which an object is pulled by several forces across a frictionless floor, as seen from overhead. In which situations does the acceleration
* x *component and (b) a *y* component? (c) In each situation, give the direction of

**To Find**

a) Which situation have x component of acceleration.

b) Which situation have y component of acceleration.

c) Direction of acceleration for each situation.

## Answer to Problem 1Q

**Solution**

a) 2, 3 and 4.

b) 1, 3 and 4.

c) 1 – Along + y-axis, 2- Along + x-axis, 3- In 4^{th} quadrant and 4- In 3^{rd} quadrant.

### Explanation of Solution

**1) Concept:**

Using the concept of net force from the Newton’s second law of motion, we can find the net force acting on the given object for given conditions.

**2) Calculations**:

a) According to Newton’s second law net force is product of mass and acceleration.

If we want x component acceleration there must be net force in x direction

So, For situation 1

Net force in x direction

**So, there is no x component of acceleration.**

For Situation 2

Net Force in x direction

**As net force is 1N, x component of acceleration is present.**

For Situation 3

Net Force in x direction

**As net force is 1N, x component of acceleration is present.**

For Situation 4

Net Force in x direction

**As net force is 1N, x component of acceleration is present.**

b)

For situation 1

Net force in y direction

**So, there is y component of acceleration.**

For Situation 2

Net Force in y direction

**As net force is no y component of acceleration is present.**

For Situation 3

Net Force in y direction

**As net force is -1N, y component of acceleration is present.**

For Situation 4

Net Force in y direction

**As net force is -4N, y component of acceleration is present.**

c) Direction of acceleration is in direction of net force.

For situation 1 there is only net force is only in +y direction so acceleration is also in +y direction.

For situation 2 there is only net force is only +x direction so acceleration is also +x direction.

For situation 3 as there is net force both in x and y direction and total net force is in fourth quadrant.

For situation 4 as there is net force both in x and y direction and total net force is in third quadrant.

**Conclusion: **Using the equations from the Newton’s second law of motion and vector algebra, it is possible to find the net force acting on the system.

### Want to see more full solutions like this?

*schedule*06:38

# Chapter 5 Solutions

Fundamentals of Physics Extended

- In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 1.91 m/s. However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice. As a result, the puck travels only one-half the distance between the players before sliding to a halt. What minimum initial speed should the puck have been given so that it reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?
*arrow_forward*2 crates are connected to each other by a rope. One crate(m=17kg) is sitting by the ledge of a cliff, and has a coefficient of friction between itself and the ground of 0.21. The second crate(m=11kg) is hanging off the edge of the cliff. Assuming that the hanging crate is heavy enough to pull the crate by the ledge to the right. Find a) and b) and Draw FBD for each crate T-E1. a) The acceleration of the system b) The tension force that connects 2 crates together Paragraph I*arrow_forward*In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 2.64 m/s. However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice. As a result, the puck travels only one-half the distance between the players before sliding to a halt. What minimum initial speed should the puck have been given so that it reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players? Number i Save for Later Units Attempts: 0 of 3 used Submit Answer*arrow_forward* - The crate, which has a mass of 100 kg, is subjected to the action of the forces. If it is originally at rest, determine the distance it slides in order to attain a speed of 6/m. The coefficient of kinetic friction between the crate and the surface is µk=0.2.
*arrow_forward*Block A with a mass of 2.25 kg rests on a tabletop. It is connected by a horizontal cord passing over a light, frictionless pulley to a hanging block B with a mass of 1.30 kg. The coefficient of kinetic friction between block A and the tabletop is 0.450. After the blocks are released from rest, determine the following: A B (a) free-body diagram of each block; (b) speed of each block after moving 3.00 cm? (Ans: 21.8 cm/s) (c) tension in the cord (Ans: 11.7 N)*arrow_forward*The diagram shows a block of mass m = 2.50 kg resting on a plane inclined at an angle of 0 = 30° to the horizontal. The coefficient of static friction between the block and the plane is Ustatic = 0.135, and the block is stationary but just on the point of sliding up the slope. 3 X E Fi D maximum magnitude of applied force = 3 N x' mg Fi The diagram shows the four forces acting on the block: an applied force F₁ acting up the slope, the block's weight mg, the normal reaction force N and the force of static friction, Ff. In this case, the force of static friction acts down the slope, opposing the tendency of the block to move up the slope. Find the the maximum magnitude of the applied force F₁ that can be exerted if the block is to remain stationary. Specify your answer by entering a number into the empty box below. 0 N.*arrow_forward* - A student, crazed by final exams, uses a force of magnitude 80 N and angle u =70 to push a 5.0 kg block across the ceiling of his room ). If the coefficient of kinetic friction between the block and the ceiling is 0.40, what is the magnitude of the block’s acceleration?
*arrow_forward*a block of mass m is held stationary on a ramp by the frictional force on it from the ramp. A force , directed up the ramp, is then applied to the block and gradually increased in magnitude from zero. During the increase, what happens to the direction and magnitude of the frictional force on the block?*arrow_forward*A mysterious force acts on all particles along a particular line and always points towards a particular point P on the line. The magnitude of the force on a particle increases as the cube of the distance from that point, that is, F∝ r3, if the distance from the P to the position of the particle is r. It has been determined that the constant of proportionality is 0.23 N/m3, i.e. the magnitude of the force on a particle can be written as 0.23r3, when the particle is at a distance r from the force center. Find the magnitude of the potential energy, in joules, of a particle subjected to this force when the particle is at a distance 0.21 m from point P assuming the potential energy to be zero when the particle is at P. PE= ?*arrow_forward* - A block of mass m1 = 3.9 kg is placed on top of a block with mass m2 = 5.4 kg. A force, F = is applied to m2, at an angle 16.1 degrees above the horizontal. If the coefficient of static friction between all moving surfaces is 0.42 and the coefficient of kinetic friction is 0.32, determine the magnitude of the minimum force that will get the blocks moving.
*arrow_forward*Question 1-b, Numerical: A 1200 kg car was moving with an initial speed of 14 m/s. The car is 60 m away from the traffic line when the traffic light turned yellow. What is the magnitude of the average friction force (Due to the breaks), measured in in kN? (1kN = 1000 N)*arrow_forward*A 15-pound box sits at rest on a horizontal surface, and there is friction between the box and the surface. One side of the surface is raised slowly to create a ramp. The friction force f opposes the direction of motion and is proportional to the normal force F exerted by the surface on the box. The proportionality constant is called the coefficient of friction, u. When the angle of the ramp. 0, reaches 25°, the box begins to slide. Find the value of u 15 pounds The value of u is (Do not round until the final answer. Then round to two decimal places as needed.)*arrow_forward*

*arrow_back_ios*

*arrow_forward_ios*

- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning