an ice skater applies a horizontal force to a 20 kg block on frictionless, level ice causing the block to accelerate uniformly at 1.4 m/s^2 to the right. after the skater stops pushing the block it slides onto a region of ice that is covered with a thin layer of sand. the coefficient of kinetic friction between the block and the sand-covered ice is 0.28 calculate the magnitude of the force applied to the block by the skater
First, let's find the force applied by the skater using Newton's second law of motion:
F = m * a
where F is the force applied, m is the mass of the block, and a is its acceleration.
F = (20 kg) * (1.4 m/s²) = 28 N
So the magnitude of the force applied to the block by the skater is 28 N.
To calculate the magnitude of the force applied to the block by the skater, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration.
Given:
Mass of the block (m) = 20 kg
Acceleration of the block (a) = 1.4 m/s²
Using Newton's second law, we can rearrange the formula to solve for force:
Force (F) = mass (m) * acceleration (a)
F = 20 kg * 1.4 m/s²
F = 28 N
So, the magnitude of the force applied to the block by the skater is 28 Newtons.
To calculate the magnitude of the force applied to the block by the skater, we need to consider the forces acting on the block.
Before the skater stops pushing the block, the only horizontal force acting on the block is the force applied by the skater. This force overcomes the inertial forces and causes the block to accelerate uniformly to the right.
Using Newton's second law (F = m*a), where F is the force, m is the mass, and a is the acceleration, we can rearrange the equation to solve for the force:
F = m * a
Here, the mass of the block (m) is given as 20 kg, and the acceleration (a) is given as 1.4 m/s^2. Plugging these numbers into the equation:
F = 20 kg * 1.4 m/s^2
= 28 N
Therefore, the magnitude of the force applied to the block by the skater is 28 Newtons.