In a twist of tradition, you pull a sled horizontally with a force of 3.3 N on a frictionless surface of snow while your dog is sitting inside the sled. If the sled accelerates at 1.27 m/s 2, and the sled’s mass is 2.8 kg, what is the mass of your dog?

To find the mass of the dog, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

Given:
Force (F) = 3.3 N
Acceleration (a) = 1.27 m/s^2
Mass of the sled (m_sled) = 2.8 kg

We need to find the mass of the dog (m_dog).

Using Newton's second law:
Force (F) = mass (m_dog) * acceleration (a)

Substituting the given values into the equation:
3.3 N = m_dog * 1.27 m/s^2

To find the mass of the dog, divide both sides of the equation by the acceleration:
m_dog = 3.3 N / 1.27 m/s^2

Calculating,
m_dog = 2.598 kg

Therefore, the mass of the dog is approximately 2.598 kg.

To find the mass of your dog, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration.

In this case, the force you are exerting on the sled is 3.3 N and the acceleration of the sled is 1.27 m/s^2. The mass of the sled is given as 2.8 kg.

So, using Newton's second law, we can rearrange the formula to solve for the mass of your dog:

Force = Mass * Acceleration

3.3 N = (2.8 kg + mass of the dog) * 1.27 m/s^2

To find the mass of your dog, we need to isolate it on one side of the equation:

3.3 N = (2.8 kg + mass of the dog) * 1.27 m/s^2

Divide both sides of the equation by 1.27 m/s^2:

3.3 N / 1.27 m/s^2 = 2.8 kg + mass of the dog

2.59 kg ≈ 2.8 kg + mass of the dog

Now, subtract 2.8 kg from both sides of the equation:

2.59 kg - 2.8 kg = mass of the dog

-0.21 kg = mass of the dog

The mass of your dog is approximately -0.21 kg. However, a negative mass does not make physical sense. Therefore, there may be an error in the calculations or the given values. Please double-check the problem and the values provided to ensure accuracy.

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