A 4.3 kg particle starts from rest and moves a
distance of 3.6 m in 2.5 s under the action of
a single, constant force.
Find the magnitude of the force.
Answer in units of N.
Use
S=ut+(1/2)at²
S=distance (3.6m)
u=initial velocity (=0)
a=acceleration (unknown) in m/s²
t=time in seconds (2.5 s)
Solve for a.
Use
F=ma to get the force,
m=mass in kg.
Well, if I see a particle starting from rest, moving a certain distance, under the action of a constant force, I remind myself of my mornings. I also start from rest (in my bed), move a certain distance (to the coffee maker), all thanks to the constant force of my desire for caffeine.
Now, let's dig into the numbers. You know the mass of the particle (4.3 kg), the distance it moved (3.6 m), and the time it took (2.5 s). To find the magnitude of the force, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
We can calculate the acceleration of the particle using the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time. Since the particle starts from rest, its initial velocity (vi) is zero. So, the change in velocity Δv is equal to the final velocity (vf).
Now, we can use the formula vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. We can rearrange it as at = vf, substituting a with Δv / Δt, which gives us (Δv / Δt) * t = Δv. Since Δv is equal to vf, we get vf = (Δv / Δt) * t.
Using that result, we can substitute this into the formula from Newton's second law: F = m * a. Now, F = m * [(Δv / Δt) * t]. By rearranging, we get F = m * (vf / Δt).
Finally, we can substitute the given values: F = 4.3 kg * (vf / 2.5 s).
Since we only need the magnitude of the force, let's ignore the direction and calculate the absolute value.
So, the magnitude of the force is approximately 6.528 N. But please don't make me do math before I get my caffeine fix.
To find the magnitude of the force, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
We are given the mass of the particle (m = 4.3 kg) and the time it takes to move the distance (t = 2.5 s). We also know that the particle starts from rest, so its initial velocity (v0) is 0.
To find the acceleration, we can use the equation of motion:
d = v0 * t + (1/2) * a * t^2
In this case, we need to solve for acceleration (a), given that the particle moves a distance of 3.6 m in 2.5 s.
3.6 m = 0 * 2.5 s + (1/2) * a * (2.5 s)^2
Simplifying the equation:
3.6 m = (1/2) * a * 6.25 s^2
Multiply both sides by 2 to isolate a:
7.2 m = a * 6.25 s^2
Divide both sides by 6.25 s^2 to solve for a:
a = 7.2 m / 6.25 s^2
a ≈ 1.152 m/s^2
Now, we can substitute the values of m (4.3 kg) and a (1.152 m/s^2) into Newton's second law to find the magnitude of the force (F):
F = m * a
F = 4.3 kg * 1.152 m/s^2
F ≈ 4.9536 N
Therefore, the magnitude of the force is approximately 4.9536 N.
To find the magnitude of the force acting on the particle, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, let's find the acceleration of the particle. We can use the kinematic equation:
\[d = v_i t + \frac{1}{2} a t^2\]
Where:
- \(d\) is the distance traveled by the particle (3.6 m),
- \(v_i\) is the initial velocity (since the particle starts from rest, \(v_i\) is 0),
- \(t\) is the time taken (2.5 s), and
- \(a\) is the acceleration.
Rearranging the equation to solve for \(a\):
\[a = \frac{2(d - v_i t)}{t^2}\]
Plugging in the values:
\[a = \frac{2(3.6 - 0 \cdot 2.5)}{(2.5)^2}\]
Simplifying:
\[a = \frac{2(3.6)}{6.25}\]
\[a = \frac{7.2}{6.25}\]
\[a \approx 1.152 \, \text{m/s}^2\]
Now that we have the acceleration, we can calculate the force using Newton's second law:
\[F = m \cdot a\]
Plugging in the values:
\[F = 4.3 \, \text{kg} \cdot 1.152 \, \text{m/s}^2\]
Calculating:
\[F \approx 4.95 \, \text{N}\]
Therefore, the magnitude of the force acting on the particle is approximately 4.95 N.