91Ó°ÊÓ

The given data is listed below as,

• The mass of the fan cart is,m=0.8 kg
• The initial velocity of the fan cart is,u=(0.9,0,0)m/s
• The force exerted by air on the fan cart is,F=(-0.3,0,0)N
• The time for which the force is exerted by air is,t=1.5s

The equation of motion mainly describes the physical system’s behavior in terms of the functions of the time.

The first equation of motion is expressed as follows,

$v-u=at$

Here,$v$ is the final velocity of the object,$u$ is the initial velocity of the object,$a$ is the acceleration and$t$ is the time taken.

The equation of the force can be expressed as:

$F=m.a$

Here,$m$is the mass of the cart, and$a$is the acceleration of the cart.

Rearrange the above equation.

$a=\frac{F}{m}$

The expression for the first equation of the motion for the fan cart is expressed as:

$$\begin{gathered} v=u+at \\ =u+\frac{F}{m}t \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} v=\left(0.9,0,0\right)\mathrm{m}/\mathrm{s}+\frac{\left(-0.3,0,0\right)\mathrm{N}}{0.8\mathrm{kg}}1.5\mathrm{s} \\ =\left(0.9,0,0\right)\mathrm{m}/\mathrm{s}+\left(-0.375,0,0\right)\mathrm{N}/\mathrm{kg}.1.5\mathrm{s} \\ =\left(0.9,0,0\right)\mathrm{m}/\mathrm{s}+\left(-0.5625,0,0\right)\mathrm{N}.\mathrm{s}/\mathrm{kg}×\frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}} \\ =\left(-0.3375,0,0\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

The equation of the change in the momentum is expressed as:

$∆p=p_2-p_1$

Here,$p_2$is the final momentum and $p_1$is the initial momentum

The above equation can be expanded as:

$$\begin{gathered} ∆p=\left(mv\right)-\left(mu\right) \\ =m\left(v-u\right) \end{gathered}$$

Substitute all the values in the above expression.

role="math" localid="1657035004902" $$\begin{gathered} ∆p=0.8kg×\left(\left(\left(0.3375,0,0\right)m/s\right)-\left(\left(0.9,0,0\right)m/s\right)\right) \\ =0.4kg×\left(\left(-0.5625,0,0\right)m/s\right) \\ =\left(-0.225,0,0\right)kg.m/s \end{gathered}$$

Thus, the change in the momentum of the fan cart is (-0.225,0,0)kg.m/s .

The equation of the change in the kinetic energy is expressed as:

$∆K.E.=K{E}_2-K{E}_1$

Here, $K{E}_2$is the final kinetic energy and $K{E}_1$is the initial kinetic energy.

The above equation can be expanded as:

$$\begin{gathered} ∆KE=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2 \\ =\frac{1}{2}m\left(v^2-u^2\right) \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} ∆K.E.=\frac{1}{2}×0.8kg×\left({\left(\left(0.3375,0,0\right)m/s\right)}^2-{\left(\left(0.9,0,0\right)m/s\right)}^2\right) \\ =0.4kg×\left(\left(0.113,0,0\right)m^2/s^2-\left(0,81,0,0\right)m^2/s^2\right) \\ =0.4kg×\left(\left(-0.697,0,0\right)m^2/s^2\right) \\ =\left(\left(-0.2788,0,0\right)kg.m^2/s^2×\frac{1J}{1kg.m^2{s}^2}\right) \\ =\left(0.2788,0,0\right)J \end{gathered}$$

Thus, the change in the kinetic energy of the fan cart is (-0.2788,0,0)J .