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?
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.
a) To calculate the acceleration of student A, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration (F = ma).
Given:
Mass of student A (m1) = 58 kg
Force applied by student A on student B (F) = 80.0 N
Using the formula F = ma, we can rearrange it to find acceleration (a):
a = F / m1
Plugging in the given values:
a = 80.0 N / 58 kg ≈ 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, we can consider the forces acting on student B.
Given:
Mass of student B (m2) = 55 kg
Acceleration of student B (a2) = 1.25 m/s^2
According to Newton's second law, the net force acting on student B is equal to the product of its mass and acceleration. This net force is also equal to the force exerted on student B by student A.
F(net) = F(A on B)
Using the formula F = ma, we can rearrange it to solve for force:
F(net) = m2 * a2
Plugging in the given values:
F(net) = 55 kg * 1.25 m/s^2 = 68.75 N
The force exerted on student B is equal in magnitude but opposite in direction to the force exerted on student A, as per Newton's third law.
Therefore, the magnitude of the force exerted by student B on the block (F(block)) is also equal to 68.75 N.
Now, we can use Newton's second law again to find the mass of the block (m(block)).
F(block) = m(block) * a(block)
Using the given value for the acceleration of student B:
68.75 N = m(block) * 1.25 m/s^2
Solving for the mass of the block:
m(block) = 68.75 N / 1.25 m/s^2 ≈ 55 kg
Therefore, the mass of the block is approximately 55 kg.
a) To calculate the acceleration of student A, we can use Newton's second law, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration. In this case, the force applied on student A is given as 80.0 N.
First, we need to calculate the net force acting on student A. According to Newton's third law, the force exerted by student A on student B is equal in magnitude but opposite in direction to the force exerted by student B on student A. Therefore, the net force acting on student A is zero, since both forces cancel each other out.
Since the net force acting on student A is zero, we can use the formula F = ma to calculate the acceleration. Rearranging the formula, we get a = F/m.
Using the given force of 80.0 N and the mass of student A, which is 58 kg, we can substitute these values into the formula to get:
a = 80.0 N / 58 kg ≈ 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 placed on student B's skateboard, we can use the same formula, F = ma, and the acceleration given as 1.25 m/s^2.
The net force acting on student B is the force exerted by student A, which we calculated as 80.0 N, minus the force required to accelerate the block.
Using the formula F = ma, we have:
80.0 N - (mass of the block × 1.25 m/s^2) = mass of student B × acceleration of student B.
We know the mass of student B is 55 kg and the acceleration of student B is given as 1.25 m/s^2.
Substituting these values into the equation, we get:
80.0 N - (mass of the block × 1.25 m/s^2) = 55 kg × 1.25 m/s^2.
Simplifying the equation, we have:
80.0 N - 1.25 m/s^2 × mass of the block = 68.75 N.
Now, we can solve for the mass of the block by rearranging the equation:
mass of the block = (80.0 N - 68.75 N) / 1.25 m/s^2.
mass of the block = 11.25 N / 1.25 m/s^2.
mass of the block ≈ 9 kg.
Therefore, the mass of the block placed on student B's skateboard is approximately 9 kg.