Question

1. a person pulls horizontally on a rope attached to a 50 kg box that is sliding away from them. the force of friction between the box and the floor is 125 n. if they wish to slow the box at a rate of 4 m/s², how hard must they pull on the rope?

for these problems, use kinematics to find the acceleration, then use newtons 2nd law to answer the questions posed.
g= -9.8 m/s² vₓ = vᵢ + at δx = vᵢt + ½at² vₓ² = vᵢ² + 2aδx δx = ½(vₓ + vᵢ)t

1. a racecar has a mass of 1000 kg. it starts from rest and travels 40.0 m in 3.0 s. the car is uniformly accelerated during the entire time. what friction force is exerted on the racecar? (use kinematics to find the acceleration, then newtons 2nd law to find the friction force)

Explanation:

Problem 3

Step1: Apply Newton's Second Law

Newton's Second Law is \(F_{net}=ma\). The net force \(F_{net}\) is the sum of the applied force \(F\) (pulling force) and the frictional force \(F_f\). Since the box is slowing down, the acceleration \(a=- 4\ m/s^{2}\) (negative because it's decelerating), \(m = 50\ kg\), and \(F_f=125\ N\). The equation becomes \(F - F_f=ma\).

Step2: Solve for the pulling force \(F\)

Rearrange the equation \(F - F_f=ma\) to \(F=ma + F_f\). Substitute \(m = 50\ kg\), \(a=-4\ m/s^{2}\), and \(F_f = 125\ N\) into the formula.

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The negative sign indicates the direction of the force (opposite to the direction of motion, which makes sense as it's a slowing - down force). The magnitude of the force is \(75\ N\).

Step1: Use the kinematic equation \(\Delta x=v_{i}t+\frac{1}{2}at^{2}\)

Since the car starts from rest, \(v_{i} = 0\ m/s\), \(\Delta x = 40.0\ m\), and \(t = 3.0\ s\). Substitute these values into the equation \(\Delta x=v_{i}t+\frac{1}{2}at^{2}\).

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Step2: Apply Newton's Second Law \(F_{net}=ma\)

Assume the force applied to the car is \(F\) (let's say from the engine) and the frictional force is \(F_f\). If we assume the only horizontal forces are the applied force (for simplicity, if we consider the net force \(F_{net}\) related to the acceleration and mass), and if we assume the force equation \(F - F_f=ma\). But if we assume that the only horizontal force causing the acceleration is related to overcoming friction (in a simple model where \(F_{net}\) is the force to overcome friction), using \(F_f = ma\) (if we assume the net force is the force to overcome friction, and the applied force is just enough to cause the acceleration considering friction). Substitute \(m = 1000\ kg\) and \(a=\frac{80}{9}\ m/s^{2}\)

$$ F_f=1000\times\frac{80}{9}=\frac{80000}{9}\approx8888.9\ N $$

Answer:

The person must pull with a force of \(75\ N\)

Problem 4