Will someone please help me with these physics questions.

Question

A baseball player hits a home run. Which best describes the action reaction pair in this system?

a. The player exerts a force on the ball and the bat accelerates its motion

b. The bat exerts a force on the ball and the ball exerts the magnitude force on the bat in the opposite direction

c. The bat exerts a force on the ball and the ball exerts a force on the air

d. The players exerts a force on the bat and the bat exerts a force on the ball

What is the kinetic friction, F_k, if a 85 kg snowboarder is moving on a flat surface of snow with the coefficient of kinetic of 0.1? 

a. 83 N

b. 850 N

c. 8.5 N

d. none of these

A student pulls a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185 N and an angle of 21.0º above the horizontal. The box has a mass of 45.0 kg and the coefficient of kinetic friction between the box and the floor is 0.27. Find the acceleration of the box.

a. 4.1 m/s²

b. 2.8 m/s²

c. 3.8 m/s²

d. 1.2 m/s²

A horizontal force of 300 N is applied to move a 55 kg sofa across a 10 m level surface. What is the net work done by this 300 N force?

a. 3000 J

b. 550 J

c. 0 J

d. 165000 J

Friction converts kinetic energy to which form?

thermal energy

total energy

potential energy

mechanical energy

Answer 1

Problem 1.

The right answer is (b). Newton's third law is about interaction of 2 bodies. In your case a bat and a ball. A bat acts on a ball; a ball acts on a bat. Those to forces equal in magnitude and opposite in direction.

Problem 2.

The right answer is (a). Kinetic friction F_k = μN. Here μ is coefficient of friction, N - normal reaction force. Because the body is on the flat surface, normal force is equal to the weight of the body. N = mg, where m - mass of the body, g - acceleration of free fall. So F_k = μmg. Now plug the numbers to get numerical value.

Problem 3.

The right answer is (d). 4 forces are acting on the box: F - force from the student at an angle Θ = 21° to the horizontal surface, weight mg vertically down, friction f against motion and normal force N vertically up. Let's chose coordinate system with x in the direction of motion and y vertically up.

The scalar form of the Newton's second law

projected on x: FcosΘ - f = ma (1)

projected on y: N + FsinΘ - mg = 0 (2)

Because f = µN and N = mg - F sinΘ (from equation (2)), we can write

F cos Θ - µmg + µFsinΘ = ma

Solving it against a, we get the answer

a = (F(cosΘ - μsinΘ) - μmg)/m

After substitution of numbers we get a = 1.2 m/s²

Problem 4

The right answer is (a). The work W = Fd, here F - applied force in the direction of motion and d - distance. Multiplying the numbers, we will get 3000J.

Problem 5

The right answer is (a). In case of friction the mechanical energy is not conserved. Mainly it is converted to thermal (internal energy). Sure it could be some energy lost to sound energy.

Hope that I answered all your questions to your satisfaction. Good luck!