Q5-72P - University Physics with Modern Physics

Question

A $6.00\text{\hspace{0.17em}kg}$ box sits on a ramp that is inclined at $37{.0}^{\circ }$ above the horizontal. The coefficient of kinetic friction between the box and the ramp is ${\mu }_k=0.30$. What horizontal force is required to move the box up the incline with a constant acceleration of  $3.60{\text{\hspace{0.17em}m/s}}^{\text{2}}$?

Step-by-Step Solution

Answer

The required horizontal force is $115\text{\hspace{0.17em}N}$.

Step 1: Identification of given data

The given data can be listed below as:

• The mass of the box is $m=6.00\text{\hspace{0.17em}kg}$.
• The angle of inclination of the box is $\theta ={37.0}^{\circ }$.
• The kinetic frictional coefficient is ${\mu }_k=0.30$ .
• The acceleration of the box is $a=3.60{\text{\hspace{0.17em}m/s}}^{\text{2}}$.

Step 2: Significance of the force

The force is described as an agent which is responsible for changing the state of motion of an object. The force exerted by an object is equal to mass and the acceleration of that object.

Step 3: Determination of the horizontal force

The free body diagram of the system has been drawn below:

According to the free body diagram, it can be identified that the summation of the force acting in the $x$ and in the $y$ direction is zero.

 

The equation of the force acting in the $x$ direction is expressed as:

 

                $F\cos \theta -{\mu }_{k}N-W\sin \theta =ma$                                                                 …(1)

 

Here,$F$ is the horizontal force,$\theta$ is the angle of inclination of the box,${\mu }_k$ is the kinetic frictional coefficient, $N$is the normal force,$W$ is the weight of the box, $m$ is the mass and is the acceleration of the box.

 

The equation of the force acting in the $x$ direction is expressed as:

$F\sin \theta +W\cos \theta =N$

Substitute the value of the above equation in the equation (i).

 $\begin{array}{c}F\cos \theta -{\mu }_k(F\sin \theta +W\cos \theta )-W\sin \theta =ma\\ F\cos \theta -{\mu }_{k}F\sin \theta +{\mu }_{k}W\cos \theta -W\sin \theta =ma\\ F(\cos \theta -{\mu }_k\sin \theta )=ma+{\mu }_{k}W\cos \theta +W\sin \theta \\ F=\frac{ma+{\mu }_{k}W\cos \theta +W\sin \theta }{(\cos \theta -{\mu }_k\sin \theta )}\end{array}$ 

Substitute $mg$ for $W$ in the above equation.

 $F=\frac{ma+{\mu }_{k}mg\cos \theta +mg\sin \theta }{(\cos \theta -{\mu }_k\sin \theta )}$

Here,$g$ is the acceleration due to gravity.

 

Substitute the values in the above equation.

 $\begin{array}{c}F=\frac{\left(6.00\text{\hspace{0.17em}kg}\right)\left(3.60{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)+\left(0.30\right)\left(6.00\text{\hspace{0.17em}kg}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\cos {37.0}^{\circ }+\left(6.00\text{\hspace{0.17em}kg}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\sin {37.0}^{\circ }}{(\cos {37.0}^{\circ }-\left(0.30\right)\sin {37.0}^{\circ })}\\ =\frac{\left(21.6{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\right)+\left(17.64{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\right)\left(0.79\right)+\left(58.8{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\right)\left(0.601\right)}{(\left(0.79\right)-\left(0.30\right)\left(0.601\right))}\\ =\frac{\left(21.6{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\right)+(13.93\text{\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}})+\left(35.33{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\right)}{(\left(0.79\right)-\left(0.1803\right))}\\ =115\text{\hspace{0.17em}N}\end{array}$

Thus, the required horizontal force is $115\text{\hspace{0.17em}N}$ .