Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 70.0 N, the second exerts a force of 92.0 N, friction is 19.0 N, and the mass of the third child plus wagon is 23.0 kg.
(b)
Calculate the acceleration (in m/s2). (Enter the magnitude.)
(c)
What would the acceleration be (in m/s2) if friction were 22.0 N? (Enter the magnitude.)
net force= m*a
92-70-19=23*a
solve for a
To calculate the acceleration in both scenarios, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the difference between the applied forces and the force of friction.
(a) First, let's find the net force in the given scenario, where friction is 19.0 N:
Net force = Sum of applied forces - Force of friction
= (70.0 N - 92.0 N) - 19.0 N
= -41.0 N
Notice that the negative sign indicates that the net force is acting in the opposite direction to the applied forces.
Next, we use Newton's second law to find the acceleration:
Net force = mass * acceleration
Rearranging the equation to solve for acceleration:
Acceleration = Net force / mass
Acceleration = -41.0 N / 23.0 kg
Acceleration ≈ -1.78 m/s^2
The negative sign indicates that the wagon is accelerating in the direction opposite to the applied forces.
(b) Now, let's consider the scenario where friction is 22.0 N:
Net force = Sum of applied forces - Force of friction
= (70.0 N - 92.0 N) - 22.0 N
= -44.0 N
Using Newton's second law again:
Acceleration = -44.0 N / 23.0 kg
Acceleration ≈ -1.91 m/s^2
In this case, the acceleration is slightly greater in magnitude compared to the previous scenario with lower friction.
To summarize:
(b) The acceleration in the first scenario, with friction of 19.0 N, is approximately -1.78 m/s^2.
(c) The acceleration in the second scenario, with friction of 22.0 N, is approximately -1.91 m/s^2.