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.
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