Question

A 4.3 kg particle starts from rest and moves a distance of 2.8 m in 3 s under the action of a single, constant force, Find the magnitude of the force?

Answer 1

the magnitude of the force acting on the particle is approximately 1.337N.

To find the magnitude of the force acting on the particle, we can use Newton's second law, which states:

F=ma

where:

F is the force,

m is the mass of the particle,

a is the acceleration of the particle.

The acceleration can be calculated using the kinematic equation:

`a=(\Delta v)/(\Delta t)`

Where:

Δv is the change in velocity,

Δt is the change in time.

The change in velocity can be determined using the kinematic equation:

`\Delta v=v_f-v_i`

where:

`v_(f)` is the final velocity,

`v_(i)` is the initial velocity.

Given that the particle starts from rest, the initial velocity
`v_(i)` is 0 m/s.

So, the force (F) can be expressed as:

`F=m\left((\Delta v)/(\Delta t)\right)`

Now, let's calculate the acceleration and then use it to find the force.

Calculate Acceleration (a):

`a=(\Delta v)/(\Delta t)`

Since the particle starts from rest 0, the final velocity is equal to the average velocity.

`\begin{aligned}& v_f=\frac{\text { distance }}{\text { time }} \\& v_f=\frac{2.8 \mathrm{~m}}{3 \mathrm{~s}} \\& v_f=0.9333 \mathrm{~m} / \mathrm{s}\end{aligned}`

Now, we can use this final velocity to calculate acceleration:

`a=(v_f-v_i)/(\Delta t)`

`\begin{aligned}& a=(v_f)/(\Delta t) \\& a=\frac{0.9333 \mathrm{~m} / \mathrm{s}}{3 \mathrm{~s}} \\& a \approx 0.3111 \mathrm{~m} / \mathrm{s}^2\end{aligned}`

Now that we have the acceleration, we can use Newton's second law:

`\begin{aligned}& F=m \cdot a \\& F=(4.3 \mathrm{~kg}) \cdot\left(0.3111 \mathrm{~m} / \mathrm{s}^2\right) \\& F \approx 1.337 \mathrm{~N}\end{aligned}`