Students are preforming an experiment about Newton's third law using skateboards. Student A has a mass of 58kg and pushes off student B with force of magnitude 80.0N. Student B had a mass of 55kg and has placed a block of unknown mass with him on his skateboard. Assume friction is negligible.

Question

a) calculate the acceleration of student A
b) if student B accelerates with a magnitude of 12.5m/s^2, what is the mass of the block?

Answer 1

a) To calculate the acceleration of student A, we can use Newton's second law:

F = ma

Where F is the force applied by student A, m is the mass of student A, and a is the acceleration. In this case, the force is given as 80.0N and the mass is 58kg. Plugging these values into the equation, we can solve for a:

80.0N = 58kg * a

a = 80.0N / 58kg

a ≈ 1.38 m/s^2

Therefore, the acceleration of student A is approximately 1.38 m/s^2.

b) To solve for the mass of the block, we can use Newton's second law again, this time to calculate the net force acting on student B:

F = ma

The force acting on student B is the force applied by student A, which we found to be 80.0N. The mass of student B is given as 55kg. The acceleration of student B is also given as 12.5 m/s^2. Plugging these values into the equation, we can solve for the mass of the block:

80.0N = (55kg + mass of block) * 12.5m/s^2

80.0N = 687.5kg*m/s^2 + 12.5mass of block

80.0N - 687.5kg*m/s^2 = 12.5mass of block

-607.5kg*m/s^2 = 12.5mass of block

mass of block = -607.5kg*m/s^2 / 12.5

mass of block = -48.6kg

Given that the mass of the block cannot have a negative value, it seems there may be an error in the calculations or the provided information. Please double-check the values and equations used.