Newton's Laws Questions: Understanding Deceleration and Resistance

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A runner decelerates from 4 m/s to 2 m/s over 8 seconds, resulting in a net force of 15 N, which includes a frictional force of 3 N. The force exerted by the wind on the runner is calculated to be 12 N after accounting for the friction. For a stone weighing 50g falling through liquid with an acceleration of 5.8 m/s², the force of resistance acting on it is determined to be 0.19 N. Both problems utilize Newton's Second Law, emphasizing the importance of identifying known values and understanding forces at play. This discussion clarifies the application of Newton's Laws in real-world scenarios.
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Right well there is this question which I am stuck on and could do with some help with! Thanks!

A runner decelerates as he runs into a windy exposed area, from a speed of 4 m/s to 2 m/s in 8 seconds. His mass is 60 kg. Assuming he also experiences a frictional force of 3 N, find the size of the force exerted by the wind on the runner.

I have found the resultant force to be 15N but to calculate the wind I'm sure it cannot just be 12N because the force pushing the man forwards has to be taken into account as well. I'm all rather confused about it!

There is also this question which I am a bit stuck with:

A stone of mass 50g is dropped into some liquid and falls vertically through it with an acceleration of 5.8 m/s^2. Find the force of resistance acting on the stone.

Many thanks!
 
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What force "pushing him forward"?

The one ambiguous point in this problem is the statement
"Assuming he also experiences a frictional force of 3 Newtons" Since there is also a reference to the wind force, I take that that is the friction of the ground on the runner. It is, I agree, ambiguous whether this force is positive or negative (whether the runner is trying to accelerat or not). Since the runner slows down, I think it is most reasonable to assert that this is also negative.

Slowing from 4 m/s to 2 m/s in 8 seconds means an acceleration of
-2/8= -0.25 m/s2. F= ma so F= (60)(-0.25)= -15 N as you calculated. Assuming that the friction force is negative, the force of the wind is -12N.

For the second problem, a stone of mass 50g (so 0.05 kg) would have acceleration -9.8 if it were falling freely. Since it is falling with acceleration -5.8 m/s2, there must be a retarding force of (0.05)(-5.8-(-9.8))= 0.05(4.1)= 0.205 N. (positive since it is upward.)
 


For both of these questions, we can use Newton's Second Law, which states that the net force on an object is equal to its mass multiplied by its acceleration (F=ma). In the first question, we are given the mass of the runner (60 kg), the initial and final velocities (4 m/s and 2 m/s), and the time (8 seconds). From this, we can calculate the deceleration of the runner using the formula a=(vf-vi)/t. Plugging in the values, we get a deceleration of 0.25 m/s^2.

Next, we can calculate the net force acting on the runner using F=ma. Since we know the mass and acceleration, we can solve for the net force, which is 15 N in this case. This includes both the frictional force and the force exerted by the wind. To find the force exerted by the wind, we can subtract the frictional force (3 N) from the net force, giving us a force of 12 N exerted by the wind.

For the second question, we have a stone with a mass of 50g (0.05 kg) falling with an acceleration of 5.8 m/s^2. Using Newton's Second Law, we can solve for the net force, which is equal to the force of resistance plus the force of gravity (weight). Since the stone is falling vertically, we can use the formula F=mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2). This gives us a weight of 0.49 N.

Now, we can plug in the values for the net force and weight into the formula F=ma, and solve for the force of resistance. This gives us a force of resistance of 0.19 N.

I hope this helps to clarify the concepts of deceleration, resistance, and Newton's Laws. Remember to always carefully read the question and identify the known values before attempting to solve for unknowns. Good luck!
 
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