91Ó°ÊÓ

It is given that,

Mass of the block is m=4.00 kg

Applied force is F =50.0 N

The normal force on the block N =13.41 N

Displacement is d =3.00 m

The problem deals with the work done which is the fundamental concept of physics. Work is the displacement of an object when force is applied to it. Also, it involves Newton’s second law of motion which can be used to find net force. Using the net force and displacement,thework done can be calculated. From the relation betweenthework done and kinetic energy,thefinal velocity of the block at 3.00m on the ramp will be found.

Formulae:

According to Newton’s second law, the force is given by,

F =ma

Work done can be expressed as,

W =Fd

$W=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$

Where, F is the force, W is the work done,$v_f,v_i$, are final and initial velocities, m is the mass, d is the displacement and a is an acceleration.

From the above figure, findthenet force to findthework done becausetheweight component is opposite to the applied force.

Normal force has been given.Usingthis, findanangle.

$$\begin{gathered} \mathrm{N}=\mathrm{mg}\mathrm{³¦´Ç²õθ} \\ 13.41\mathrm{N}=4.00\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2×\mathrm{³¦´Ç²õθ} \\ \mathrm{³¦´Ç²õθ}=\frac{13.41\mathrm{N}}{4.00\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2} \\ =0.3420 \\ \mathrm{θ}={\cos }^{-1}\left(0.3420\right) \\ =69.995° \end{gathered}$$

So, find another component of weight,

$$\begin{gathered} \mathrm{mg}\sin \mathrm{θ}=4.00\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2×\sin \left(69.995\right) \\ \mathrm{mg}\sin \mathrm{θ}=36.8347\mathrm{N} \end{gathered}$$

Using Newton’s second law, findthenet force,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=\mathrm{F}-\mathrm{mgsin}\mathrm{θ} \\ =50\mathrm{N}-36.8347\mathrm{N} \\ =13.1652\mathrm{N} \end{gathered}$$

Now, find the work done by this net force,

$W=F_{ne{t}_{}}d\cos \mathrm{Ï•}$

Where$\mathrm{Ï•}$Is the angle between force and displacement.

Here, the angle $\mathrm{ϕ}=0°$,

$$\begin{gathered} \mathrm{W}=\left(13.1652\mathrm{N}\right)\left(3.00\mathrm{m}\right)\cos \left(0\right) \\ =39.4956\mathrm{J} \end{gathered}$$

The other equation is known, which relates the work done and kinetic energy,

$W=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$

The block is pulled from rest, so,$v_i=0$

$$\begin{gathered} 39.4956\mathrm{J}=\frac{1}{2}\left(4.00\mathrm{kg}\right){\mathrm{v}}_{\mathrm{f}}^2-\frac{1}{2}\left(4.00\mathrm{kg}\right){\left(0\right)}^2 \\ {\mathrm{v}}_{\mathrm{f}}^2=\frac{\left(39.4956\mathrm{J}\right)×2}{\left(4.00\mathrm{kg}\right)} \\ =19.7478{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\mathrm{v}}_{\mathrm{f}}^{}=4.44\mathrm{m}/\mathrm{s} \end{gathered}$$