Question

A 2.30 kg box is moving to the right with speed 9.00 m/s on a horizontal, frictionless surface. at t = 0 a horizontal force is applied to the box. the force is directed to the left and has magnitude f(t)=( 6.00 n/s2 )t2

Answer 1

The question concerns calculating the effects of a time-varying horizontal force on a moving box's acceleration, velocity, and displacement, using Newton's second law in a frictionless environment.

Step-by-step explanation:

The student's question pertains to a Newtonian mechanics problem where a force varying with time (f(t)=(6.00 N/s²)t²) is applied to a box moving on a frictionless surface. The mass of the box is given as 2.30 kg, and it initially moves to the right with a speed of 9.00 m/s. The student is likely interested in determining the resulting acceleration, velocity at a certain time, or the displacement of the box after the force has been applied for a specific duration.

To solve such a problem, we would use Newton's second law, F=ma, where F is the force, m is the mass, and a is the acceleration. Since the force is given as a function of time, the acceleration can also be expressed as a function of time. From there, one can integrate to find velocity and displacement as functions of time.

Answer 2

The distance traveled before the speed is reduced to zero is
`C_(2)=0`. If the force continues to be applied, the speed of the box at t=3.00s is 54m/s.

To solve this problem, we can use the kinematic equations of motion and Newton's second law. Let's break it down into parts:

Part (a):

The net force acting on the box is given by Newton's second law:

`F_{\text {net }}=m \cdot a`

The acceleration a is the second derivative of position x with respect to time t:

`a(t)=(d^2 x)/(d t^2)`

The force F(t) is given by the problem as
`F(t)=\left(6.00 \mathrm{~m} / \mathrm{s}^2\right) t^2`. We can find acceleration by taking the derivative twice:

`a(t)=(d^2 x)/(d t^2)=(d)/(d t)\left(6.00 t^2\right)`

Integrate this acceleration function to get the velocity function, and then integrate the velocity function to get the position function:

`\begin{aligned}& v(t)=\int a(t) d t \\& x(t)=\int v(t) d t\end{aligned}`

To find the distance the box moves before its speed is reduced to zero, you need to find the time t when v(t)=0. Solve for t in the velocity equation and substitute this into the position equation to find the distance.

Acceleration (a(t)):

`a(t)=(d^2 x)/(d t^2)=(d)/(d t)\left(6.00 t^2\right)=12.00 t`

Velocity (v(t)):

`v(t)=\int a(t) d t=\int 12.00 t d t=6.00 t^2+C_1`

where
`C_(1)` is an integration constant.

Position (x(t)):

`x(t)=\int v(t) d t=\int\left(6.00 t^2+C_1\right) d t=2.00 t^3+C_1 t+C_2`

where
`C_(1)` and
`C_(2)` are integration constants.

Now, to find the distance the box moves before its speed is reduced to zero, we set v(t)=0 and solve for t:

`6.00 t^2+C_1=0`

Solving for t gives t=0 (ignoring the negative solution, as we are dealing with time).

Substitute this t value into the position equation:

`\begin{aligned}& x(t)=2.00 t^3+C_1 t+C_2 \\& x(0)=C_2\end{aligned}`

So, the distance the box moves before its speed is reduced to zero is given by
`C_(2)`.

Given
`v(t)=6.00 t^2+C_1, \text { and } x(t)=2.00 t^3+C_1 t+C_2`, we know that v(0)=0 (initial speed is given as 9 m/s). Substituting t=0 into v(t), we find
`C_(1)=0`.

Now,
`x(t)=2.00 t^3+C_2`. The distance before the speed is reduced to zero is given by
`x(0)=C_2`.

`x(0)=C_2`

So,
`C_(2)` is the distance traveled before the speed is reduced to zero.

`\begin{aligned}& C_2=x(0) \\& C_2=2.00 *(0)^3+C_2 \\& C_2=0\end{aligned}`

So, the distance traveled before the speed is reduced to zero is
`C_(2)=0`.

Part (b):

To find the speed at t=3.00s, use the velocity function v(t) that you obtained in part (a) and evaluate it at t=3.00s.

`v(3.00)=6.00(3.00)^2+C_1`

`\begin{aligned}& v(3.00)=6.00 * 9+C_1 \\& v(3.00)=54+C_1\end{aligned}`

`\begin{aligned}& v(3.00)=6.00 * 9+C_1 \\& v(3.00)=54+0\end{aligned}`

Question:

A 2.00-kg box is moving to the right with speed 9 m/s

on a horizontal, frictionless surface. At t=0 a horizontal

force is applied to the box. The force is directed to the left and

has magnitude
`F(t)=(6.00 m/s^2)t^2` .

(a) What distance does the box move from its position at t=0 before its speed is reduced to zero?

(b) If the force continues to be applied, what is the speed of the box at t=3.00s.