Q 31E - University Physics with Modern Physics

Question

Question: A box with mass 10.0 kg moves on a ramp that is inclined at an angle of above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is . Calculate the magnitude of the acceleration of the box if you push on the box with a constant force that is parallel to the ramp surface and (a) directed down the ramp, moving the box down the ramp; (b) directed up the ramp, moving the box up the ramp.

Step-by-Step Solution

Answer

a) The magnitude of the acceleration directed down the ramp is, $18.314m/s^2$.

b) The magnitude of the acceleration directed up the ramp is, $2.286m/s^2$. 

Step 1: Identification of given data

The given data can be listed below as,

• The mass of the box is, $m=10.0 \text{kg}$. 
• The angle of inclination of ramp is, $\theta =55°$.
• The coefficient of kinetic friction is, ${\mu }_k=0.300$.
• The constant force on the box is, $F=120.0 \text{N}$.

Step 2: Significance of magnitude of acceleration

The rate of change of velocity with respect to time is known as acceleration.  Magnitude of acceleration can be define as rate of change in the magnitude of velocity and rate of changing the direction of motion.

Step 3: Determination of magnitude of the acceleration directed down the ramp.

The expression for the normal force along y-axis can be expressed as,

 $N=m g \cos \theta$

Here m is the mass, and g is the acceleration due to gravity, $\theta$ is the angle of inclination.

 

Substitute 10.0kg for m, and $9.8m/s^2$ for g, and $55°$ for $\theta$ in the above equation.

 $\begin{array}{rcl}N& =& 10.0 {\text{kg×9.8m/s}}^2\cos \left(55°\right)\\ & =& 56.21 \text{N}\end{array}$

The expression for the frictional force can be expressed as,

$F_r={\mu }_k N$ 

Here ${\mu }_k$ is the coefficient of kinetic friction, N is the normal force.

Substitute 0.300 for ${\mu }_k$, and 56.21N for N in the above equation.

 $\begin{array}{rcl}{F}_r& =& 0.300\times 56.21 \text{N}\\ & =& 16.863 \text{N}\end{array}$

The expression for the Newton’s second  law can be expressed as,

$$\begin{gathered} \sum F_x=m a \\ m a=F+ m g \sin \theta -F_r \\ a=\frac{F+ m g \sin \theta -F_r}{m} \end{gathered}$$

Here a is the acceleration of the box.

 

Substitute 10.0 kg for m, $9.8m/s^2$ for g, 120.0 N for F, and $55°$ for $\theta$, 16.863N for $F_r$ in the above equation.

$\begin{array}{rcl}a& =& \frac{{\displaystyle 120 \text{N} +\text{10} \text{kg}\times \text{9.8} m/s^2\times \sin (55°) -16.863 \text{N}}}{10kg}\\ & =& 18.34 m/s^2\end{array}$

Hence, required magnitude of acceleration directed down the ramp is $\begin{array}{rcl}& & 18.34 m/s^2\end{array}$.

Step 4: Determination of magnitude of acceleration directed up the ramp.

The expression for the Newton’s second law can be expressed as,

 $$\begin{gathered} \sum F_x=m a \\ m a=F- m g \sin \theta -F_r \\ a=\frac{F- m g \sin \theta -F_r}{m} \end{gathered}$$

Substitute 10.0 kg for m, $9.8m/s^2$ for g, 120.0 N for F, and $55°$ for $\theta$, 16.863N for $F_r$ in the above equation.       

 $\begin{array}{rcl}a& =& \frac{{\displaystyle 120 \text{N} +\text{10} \text{kg}\times \text{9.8} m/s^2\times \sin (55°) -16.863 \text{N}}}{10kg}\\ & =& 2.286 m/s^2\end{array}$

Hence, required magnitude of acceleration directed up the lamp is, $\begin{array}{rcl}& & 2.286 m/s^2\end{array}$.