A 1.5 kg chunk of ice slides down a roof sloped at a 60 degree angle with the horizontal.

If there is no friction, determine the acceleration of the ice.
Show that the amount of mass doesn't matter in part a.

To determine the acceleration of the ice sliding down a roof with no friction, we can use Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration.

First, let's find the component of the gravitational force acting on the ice parallel to the roof. We can do this by multiplying the gravitational force (weight) of the ice by the sine of the angle between the roof and the horizontal.

The weight of the ice can be calculated using the formula:

Weight = mass × gravity

where the mass is given as 1.5 kg and gravity is approximately 9.8 m/s^2.

Weight = 1.5 kg × 9.8 m/s^2
Weight = 14.7 N

The component of the weight parallel to the roof is given by:

Force_parallel = Weight × sin(angle)

Substituting the given angle of 60 degrees:

Force_parallel = 14.7 N × sin(60 degrees)
Force_parallel = 14.7 N × 0.866
Force_parallel ≈ 12.73 N

Since there is no friction, this force is the only force acting on the ice in the direction of motion. Applying Newton's second law:

Force_parallel = mass × acceleration

Rearranging the equation and plugging in the values:

acceleration = Force_parallel / mass
acceleration = 12.73 N / 1.5 kg
acceleration ≈ 8.49 m/s^2

So, the acceleration of the ice sliding down the roof with no friction is approximately 8.49 m/s^2.

Now, let's address part b - showing that the amount of mass doesn't matter in part a. As we can see from the equation:

acceleration = Force_parallel / mass

The mass is directly proportional to the weight, as weight is calculated by multiplying mass by gravity. Therefore, even if we change the mass of the ice, the weight and the force parallel to the roof will change proportionally. Consequently, the mass cancels out when calculating the acceleration, and the acceleration remains the same regardless of the mass.