A man of mass 75 kg and a women of mass 55 kg stand facing each other on an ice rink , both wearing ice skates . The women pushes the man with a horizontal force of 85 N in the positive x direction . a) What is the man s acceleration? b) What is the reaction force acting on the women? c) Calculate the women s acceleration.
Third law, action is equal and opposite to reaction, same |F| of 85 Newtons on both, positive on man, negative on woman.
F = m a
a of man = 85/75
a of woman = -85/55
a car mass 100kg is moving up on incline of 15 degrees with horizontal.the enginr of the car applies a force of 5000n the confficient of kinetic friction between the surface of th tyic and road is 0,2.draw a free body diagram showing all the forces actin on the car
a) Well, let's calculate the man's acceleration. Remember, we can use Newton's second law: F = ma. The force applied by the woman is 85 N, and the man's mass is 75 kg. So, the formula becomes 85 N = 75 kg * a. Solving for a, we get a value of 1.13 m/s². So, the man's acceleration is 1.13 m/s².
b) Ah, the reaction force on the woman. Well, Newton's third law states that for every action, there is an equal and opposite reaction. So, the reaction force on the woman will be equal in magnitude but opposite in direction to the force she applied on the man. Therefore, the reaction force on the woman is also 85 N but in the negative x direction.
c) Now, let's find the woman's acceleration. Since we know the reaction force acting on her is 85 N and her mass is 55 kg, we can use the same formula as before: F = ma. Plugging in the values, we get 85 N = 55 kg * a. Solving for a gives us an acceleration of 1.55 m/s². So, the woman's acceleration is 1.55 m/s².
Just remember, when it comes to relationships, it's all about giving each other a push in the right direction!
To find the answers to these questions, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration: F_net = m * a.
a) To calculate the man's acceleration, we can first find the net force acting on him. The force applied by the woman is 85 N in the positive x direction. Since there are no other forces mentioned, this is the net force acting on the man. Therefore, we have:
F_net = 85 N
m_man = 75 kg
Using Newton's second law, we can rearrange the equation to solve for acceleration:
F_net = m_man * a_man
85 N = 75 kg * a_man
Dividing both sides of the equation by 75 kg, we can find the man's acceleration:
a_man = 85 N / 75 kg
a_man ≈ 1.13 m/s^2
b) The reaction force acting on the woman is equal in magnitude but opposite in direction to the force she applied on the man. Therefore, the reaction force will also be 85 N, but in the negative x direction.
c) To calculate the women's acceleration, we can use the same equation we used to find the man's acceleration:
F_net = m_woman * a_woman
Since the reaction force is acting on the woman in the negative x direction, we need to consider the negative sign:
-85 N = 55 kg * a_woman
Dividing both sides of the equation by 55 kg, we can find the woman's acceleration:
a_woman = -85 N / 55 kg
a_woman ≈ -1.55 m/s^2
Therefore, the woman's acceleration is approximately -1.55 m/s^2, indicating that she moves in the negative x direction.
Post you questions separately,. not as a reposnse to someone else's question. We cannot provide diagrams for you in any case.
Finally, the coefficient of kinetic friction does not apply to a object (like a car tire) that is rolling without slipping. Perhaps the instructor is not aware of that.
The "applied force", plus the weight, is the force exerted by the road on the tires.