Isabella pushes a box of mass 5.25 kilograms with a force of 15.7 newtons, it accelerates by a rate of 2.50 meters/second². What is the force due to friction in this scenario?
Understand the Problem
The question is asking for the force acting on a box being pushed, based on its mass, the applied force, and the acceleration. This involves using the principles of Newton's Second Law of Motion to determine the net force acting on the object.
Answer
The force due to friction is $-2.60 \text{ N}$.
Answer for screen readers
The force due to friction is approximately $-2.60 \text{ N}$.
Steps to Solve
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Identify Given Values We have the following information:
- Mass of the box, $m = 5.25 \text{ kg}$
- Applied force, $F_{applied} = 15.7 \text{ N}$
- Acceleration, $a = 2.50 \text{ m/s}^2$
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Calculate the Force of Friction We can use Newton's Second Law of Motion which states that the net force is equal to mass times acceleration. The equation for net force is:
$$ F_{net} = m \cdot a $$
Substituting in the values we have:
$$ F_{net} = 5.25 \text{ kg} \cdot 2.50 \text{ m/s}^2 $$
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Calculate the Value of $F_{net}$ Now we perform the calculation:
$$ F_{net} = 5.25 \cdot 2.50 = 13.125 \text{ N} $$
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Determine the Force of Friction The net force can also be calculated as the applied force minus the force of friction. Thus, we can rearrange the equation:
$$ F_{friction} = F_{applied} - F_{net} $$
Substituting in the values:
$$ F_{friction} = 15.7 \text{ N} - 13.125 \text{ N} $$
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Calculate the Force of Friction Now we perform this final calculation:
$$ F_{friction} = 15.7 - 13.125 = 2.575 \text{ N} $$
Since we are asked for the force due to friction, we will report this as negative because it opposes the direction of the applied force:
$$ F_{friction} = -2.575 \text{ N} $$
The force due to friction is approximately $-2.60 \text{ N}$.
More Information
In this scenario, the negative sign indicates that the force of friction acts in the opposite direction of the applied force. This situation is a practical application of Newton's Second Law and helps us understand the interaction between applied forces and resistive forces like friction.
Tips
- Forgetting to convert units when necessary.
- Not applying the negative sign for forces opposing the direction of motion.
- Miscalculating the net force and force of friction due to arithmetic errors.
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