Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 95.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 21.0 kg.

(c) Calculate the acceleration.
m/s2
(d) What would the acceleration be if friction is 20.0 N?

c. a = (95-75-12)/21 = 0.381 m/s^2.

d. a = (95-75-20)/21 = 0/21 = 0.

(c) Well, first let's add up the forces. The net force is the sum of the two pushing forces minus the force of friction. So the net force is (75.0 N - 95.0 N) - 12.0 N = -32.0 N.

To calculate the acceleration, we use Newton's second law, F = ma. In this case, the net force is -32.0 N and the mass is 21.0 kg. So, -32.0 N = 21.0 kg * a.

Now let's solve for a. Dividing both sides by 21.0 kg gives us a = -1.52 m/s^2.

Wait, negative acceleration? That means the child and wagon are slowing down!

(d) If the friction is increased to 20.0 N, let's calculate the new net force. The net force would now be (75.0 N - 95.0 N) - 20.0 N = -40.0 N.

We can plug this net force into our formula for Newton's second law, F = ma. So, -40.0 N = 21.0 kg * a.

Dividing both sides by 21.0 kg gives us a = -1.90 m/s^2.

Looks like the child and wagon will slow down even more with the increased friction. Difficult situation for the kid!

To calculate the acceleration of the third child plus wagon, we need to determine the net force acting on them.

(a) Calculate the net force:
Net force = Sum of all forces
= Force exerted by the first child + Force exerted by the second child - Frictional force
= 75.0 N + 95.0 N - 12.0 N
= 158.0 N - 12.0 N
= 146.0 N

(b) Calculate the acceleration:
Using Newton's second law, F = ma, where F is the net force and m is the mass:
Net force (F) = mass (m) × acceleration (a)
146.0 N = 21.0 kg × a

Solving for a:
a = 146.0 N / 21.0 kg
a ≈ 6.9524 m/s²

Therefore, the acceleration of the third child plus wagon is approximately 6.9524 m/s².

(c) If friction is 20.0 N, we need to recalculate the net force:

Net force = Force exerted by the first child + Force exerted by the second child - Frictional force
= 75.0 N + 95.0 N - 20.0 N
= 170.0 N - 20.0 N
= 150.0 N

Recalculating the acceleration:
150.0 N = 21.0 kg × a

Solving for a:
a = 150.0 N / 21.0 kg
a ≈ 7.1429 m/s²

Therefore, if friction is 20.0 N, the acceleration of the third child plus wagon would be approximately 7.1429 m/s².

To calculate the acceleration of the third child plus wagon, we can use Newton's second law of motion, 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) Calculate the net force acting on the third child plus wagon:
Net force = total force - frictional force
= (75.0 N - 95.0 N) - 12.0 N
= -32.0 N

Keep in mind that the negative sign indicates that the net force is in the opposite direction of the applied forces.

(b) Calculate the acceleration using Newton's second law formula:
Acceleration = Net force / mass
= (-32.0 N) / (21.0 kg)
≈ -1.52 m/s²

The negative sign means that the acceleration is in the opposite direction of the applied forces.

(c) Now, let's calculate the acceleration when the friction is 20.0 N:
Net force = total force - frictional force
= (75.0 N - 95.0 N) - 20.0 N
= -40.0 N

Acceleration = Net force / mass
= (-40.0 N) / (21.0 kg)
≈ -1.90 m/s²

Again, the negative sign indicates that the acceleration is in the opposite direction of the applied forces.