If a dog chases its tail in a perfect circle of radius r at constant speed v, which of the following is true?

A. The dog's acceleration is zero.
B. The dog's velocity is constant.
C. The sum of the forces, F , acting on the dog is zero.
D. The time it takes the dog to make one circle is given by t = 2 r/v.

You attempt to push a heavy box of mass M sitting at rest on a level floor, but are unable to move it with a force F . If the coefficients of static and kinetic friction between the box and the floor are s and k respectively, and s > k, what can we definitely say about the force F ? (g = gravitational acceleration)

A. F < M g
B. F < kM g
C. F < sM g
D. 2F > sM g

Right answer choices for B-D

B. F < µkMg
C. F < µsMg
D. 2F > µsMg

To answer the first question about the dog chasing its tail, we can use some principles from physics. Let's break down the options:

A. The dog's acceleration is zero: In this case, the dog would be moving at a constant speed in a perfect circle, which means its velocity would be changing direction. Because velocity is a vector quantity, a change in direction implies acceleration. So, option A is not true.

B. The dog's velocity is constant: The dog is moving in a perfect circle, which means its direction is changing continuously. Therefore, its velocity is not constant. So, option B is not true.

C. The sum of the forces, F, acting on the dog is zero: The dog is constantly changing direction, which requires a centripetal force pointing towards the center of the circle. This force is provided by the tension in the dog's body. Since there is a force acting on the dog, the sum of forces is not zero. So, option C is not true.

D. The time it takes the dog to make one circle is given by t = 2r/v: This is the correct answer. The time it takes for an object to complete a full circle is given by the formula t = (2πr)/v, where r is the radius of the circle and v is the speed of the dog. Note that π is a constant value. Therefore, option D is true.

Now, let's move on to the second question about the heavy box and the force required to move it.

In this case, we are given that the force you apply to the box, F, is not enough to move it. We also know that the coefficients of static and kinetic friction between the box and the floor are s and k, respectively, with s > k.

To determine the relationship between F and M g (where M is the mass of the box and g is the gravitational acceleration), we need to consider the maximum value of the static friction force, which is given by F_s = sMg.

Since s > k, the maximum static friction is greater than the kinetic friction force, which is given by F_k = kMg.

To overcome the static friction and start moving the box, the force you apply, F, must be greater than or equal to the maximum static friction, F_s. So, we can definitely say that:

F ≥ F_s = sMg

Therefore, option C, F < sMg, is not true.

On the other hand, the force you apply, F, is less than the maximum kinetic friction, F_k = kMg, because it is not enough to move the box. So, we can definitely say that:

F < F_k = kMg

Therefore, option B, F < kMg, is true.

The correct answer to the second question is:

B. F < kMg