91Ó°ÊÓ

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,$\mathrm{θ}=55°$.
• The coefficient of kinetic friction is, ${\mathrm{μ}}_k=0.300$.
• The constant force on the box is, $F=120.0\text{N}$.

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.

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

$N=mg\cos \mathrm{θ}$

Here m is the mass, and g is the acceleration due to gravity,$\mathrm{θ}$is the angle of inclination.

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

$\begin{array}{rcl}N& =& 10.0{\text{°ì²µÃ—9.8³¾/²õ}}^2\cos \left(55°\right)\\ & =& 56.21\text{N}\end{array}$

The expression for the frictional force can be expressed as,

$F_r={\mathrm{μ}}_{k}N$

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

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

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

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

$$\begin{gathered} ∑F_x=ma \\ ma=F+mg\sin \mathrm{θ}-F_r \\ a=\frac{F+mg\sin \mathrm{θ}-F_r}{m} \end{gathered}$$

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 $\mathrm{θ}$, 16.863N for$F_r$ in the above equation.

role="math" localid="1668409489310" $\begin{array}{rcl}a& =& \frac{{\displaystyle 120\text{N}+\text{10}\text{kg}×\text{9.8}m/s^2×\sin (55°)-16.863\text{N}}}{10kg}\\ & =& 18.34m/s^2\end{array}$

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

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

$$\begin{gathered} ∑F_x=ma \\ ma=F-mg\sin \mathrm{θ}-F_r \\ a=\frac{F-mg\sin \mathrm{θ}-F_r}{m} \end{gathered}$$

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

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

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