Students are performing an experiment about Newton’s third law using skateboards. Student A has a mass of 58 kg and pushes off student B with a force of magnitude 80.0 N. Student B has a mass of 55 kg and has placed a block of unknown mass with him on his skateboard. Assume friction is negligible.
a) Calculate the acceleration of student A .
b) If student B accelerates with a magnitude of 1.25 m/s^2, what is the mass of the block?
a) To calculate the acceleration of student A, we can use Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration.
The force exerted by student A on student B is equal to the force exerted by student B on student A (Newton's third law). Therefore, the magnitude of the force exerted on student A is also 80.0 N.
Using the formula F = ma, where F is the force, m is the mass, and a is the acceleration, we can rearrange the formula to solve for acceleration: a = F/m.
Plugging in the values, we have a = 80.0 N / 58 kg = 1.38 m/s^2.
b) To find the mass of the block, we can again use Newton's second law. The force exerted on student B is the force exerted by student A, which is 80.0 N. The mass of student B plus the mass of the block is 55 kg + m (mass of the block).
Using the formula F = ma, where F is the force, m is the mass, and a is the acceleration, we can rearrange the formula to solve for the mass of the block: m = (F - ma) / a.
Plugging in the values, we have m = (80.0 N - 55 kg * 1.25 m/s^2) / 1.25 m/s^2.
Simplifying, we have m = (80.0 N - 68.75) / 1.25 m/s^2.
Calculating further, we have m = 11.25 / 1.25 = 9 kg.
Therefore, the mass of the block is 9 kg.
To calculate the acceleration of student A, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
a) First, let's calculate the force on student A. The force is equal to the mass of student A multiplied by the acceleration of student A.
Force = mass × acceleration
Force = 58 kg × acceleration
We know that the force applied by student A on student B is 80.0 N. So,
80.0 N = 58 kg × acceleration
Now, we can solve for the acceleration:
acceleration = 80.0 N / 58 kg
acceleration ≈ 1.38 m/s^2
Therefore, the acceleration of student A is approximately 1.38 m/s^2.
b) To find the mass of the block on student B's skateboard, we can use the concept of Newton's second law again. The force on student B is due to the interaction with the block on the skateboard.
Force = mass × acceleration
Force = (mass of student B + mass of block) × acceleration
We know that the force on student B is 55 kg × 1.25 m/s^2 (as given in the question). So,
55 kg × 1.25 m/s^2 = (mass of student B + mass of block) × 1.25 m/s^2
Dividing both sides by 1.25 m/s^2, we get:
55 kg = mass of student B + mass of block
Given that the mass of student B is 55 kg, the mass of the block can be calculated as:
mass of block = 55 kg - mass of student B
Substituting the value of mass of student B as 55 kg, we get:
mass of block = 55 kg - 55 kg
mass of block = 0 kg
Therefore, the mass of the block on student B's skateboard is 0 kg.
To solve this problem, we'll use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In this case, when student A pushes off student B with a force, student B also exerts an equal and opposite force on student A.
a) To calculate the acceleration of student A, we'll use Newton's second law of motion, which states that the force exerted on an object is equal to its mass multiplied by its acceleration (F = ma).
The force exerted on student A by student B is equal to the force applied by student A on student B. Therefore, we have:
Force on student A = Force on student B
Given:
Mass of student A (m₁) = 58 kg
Force on student B (F) = 80.0 N
Applying Newton's second law of motion,
Force on student A = m₁ * acceleration of student A
80 N = 58 kg * acceleration of student A
Rearranging the equation, we can find the acceleration of student A:
acceleration of student A = 80 N / 58 kg
acceleration of student A ≈ 1.3793 m/s^2
Therefore, the acceleration of student A is approximately 1.3793 m/s^2.
b) To calculate the mass of the block (m₂) on student B's skateboard, we can use the same principle. The net force on student B is equal to the force applied by student A minus the force due to the block (F_net = F - F_block).
Given:
Mass of student B (m_b) = 55 kg
Acceleration of student B (a_b) = 1.25 m/s^2
Applying Newton's second law of motion to student B,
F_net = m_b * a_b
F - F_block = 55 kg * 1.25 m/s^2
Since we know the mass of student B (m_b), we can rewrite the equation as:
Force on student A - Force due to the block = 55 kg * 1.25 m/s^2
Substituting the value of the force on student A (80 N) from part a),
80 N - F_block = 55 kg * 1.25 m/s^2
Rearranging the equation, we can find the force due to the block (F_block):
F_block = 80 N - 55 kg * 1.25 m/s^2
Now, we have F_block and can apply Newton's second law of motion to find the mass of the block (m₂):
F_block = m₂ * a_b
80 N - 55 kg * 1.25 m/s^2 = m₂ * 1.25 m/s^2
Simplifying the equation,
80 N - 55 kg * 1.25 m/s^2 = 1.25 m/s^2 * m₂
m₂ = (80 N - 55 kg * 1.25 m/s^2) / 1.25 m/s^2
Calculating this expression will give you the mass of the block (m₂).