Question

A 4.00 kg block is pulled up a frictionless inclined plane by a 50.0 N force that is parallel to the plane, starting from rest. The normal force on the block from the plane has magnitude 13.41 N. What is the block’s speed when its displacement up the ramp is 3.00 m?

Short answer

Block’s speed at 3.00 m on the ramp is 4.44 m/s.

Step-by-step solution

Given information

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

Determining the concept of work done

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.

Determining the block’s speed at 3.00 on the ramp

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

Normal force has been given.Usingthis, findanangle.

$$\begin{gathered} \mathrm{N}=\mathrm{mg}\mathrm{cos\theta } \\ 13.41\mathrm{N}=4.00\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \mathrm{cos\theta } \\ \mathrm{cos\theta }=\frac{13.41\mathrm{N}}{4.00\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2} \\ =0.3420 \\ \mathrm{\theta }={\cos }^{-1}\left(0.3420\right) \\ =69.995° \end{gathered}$$

So, find another component of weight,

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

Using Newton’s second law, findthenet force,

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

Now, find the work done by this net force,

$W=F_{ne{t}_{}}d\cos \varphi$

Where$\varphi$Is the angle between force and displacement.

Here, the angle $\varphi =0°$,

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

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,$v_i=0$

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

Hence, the speed of the block at 3.00m on the ramp is ${\mathrm{v}}_{\mathrm{f}}^{}=4.44\mathrm{m}/\mathrm{s}$.