To solve this problem, we can use Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
a) To calculate the acceleration of student A, we can use the equation:
F = m * a
Where F is the net force, m is the mass, and a is the acceleration.
Given:
Force exerted by student A on student B (net force) = 80.0 N
Mass of student A = 58 kg
Using the equation, we can rearrange it to solve for acceleration:
a = F / m
a = 80.0 N / 58 kg
a ≈ 1.38 m/s^2
Therefore, the acceleration of student A is approximately 1.38 m/s^2.
b) To calculate the mass of the block, we can use the same equation:
F = m * a
Where F is the net force, m is the mass, and a is the acceleration.
Given:
Acceleration of student B = 1.25 m/s^2
Mass of student B = 55 kg
The net force acting on student B is due to the force exerted by student A.
Using the equation, we can rearrange it to solve for mass:
m * a = F
(mass of student B + mass of the block) * 1.25 m/s^2 = 80.0 N
(55 kg + mass of the block) * 1.25 m/s^2 = 80.0 N
Simplifying the equation:
68.75 kg + 1.25 * mass of the block = 80.0 N
Rearranging the equation to solve for the mass of the block:
1.25 * mass of the block = 80.0 N - 68.75 kg
1.25 * mass of the block = 11.25 kg
Dividing both sides by 1.25 to solve for the mass of the block:
mass of the block = 11.25 kg / 1.25
mass of the block = 9 kg
Therefore, the mass of the block is 9 kg.
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?
1 answer
1 answer