Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the man is 88 kg, and the mass of the woman is 58 kg. The woman pushes on the man with a force of 56 N due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman.
To determine the acceleration of each skater, we can apply Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
Let's start with the man:
(a) Man's acceleration:
1. Identify the given values:
- Mass of the man (m1) = 88 kg
- Force applied by the woman (F) = 56 N (due east)
2. Determine the acceleration of the man using Newton's second law (F = ma), where F is the net force and a is the acceleration:
- Rearrange the formula to solve for acceleration: a = F / m1
- Substitute the known values: a = 56 N / 88 kg
- Calculate the acceleration: a ≈ 0.636 m/s² (due east)
The man's acceleration is approximately 0.636 m/s², and it is directed eastward.
Now let's find the woman's acceleration:
(b) Woman's acceleration:
1. Identify the given values:
- Mass of the woman (m2) = 58 kg
- Force applied by the woman (F) = 56 N (due east)
2. Determine the acceleration of the woman using Newton's second law (F = ma), where F is the net force and a is the acceleration:
- Rearrange the formula to solve for acceleration: a = F / m2
- Substitute the known values: a = 56 N / 58 kg
- Calculate the acceleration: a ≈ 0.966 m/s² (due east)
The woman's acceleration is approximately 0.966 m/s², and it is directed eastward as well.