91Ó°ÊÓ

The mass of the box is,$m=5.00kg$.

Newton’s second law states that force on the body is equal to the product of mass and acceleration.

The free-body diagram has to be drawn. From the given figure, the acceleration of the box by using its expression can be found. The magnitude of normal force on the box from the ramp can be found by using Newton’s second law.

Formulae:

$$\begin{gathered} \stackrel{→}{F}net=m\stackrel{→}{a} \\ a=\frac{v_f-v_i}{t_2-t_1} \end{gathered}$$

From the figure, initial horizontal velocity of the box is $v_i=4m/sat{t}_1=0s$and final velocity is$v_f=-1m/sat{t}_2=2s.$

By using the expression of acceleration as,

$$\begin{gathered} a=\frac{-1m/s-4m/s}{2s-0s} \\ =-2.50m/s^2 \end{gathered}$$

By applying Newton’s second law along x axis as,

$\stackrel{→}{F}net=m\stackrel{→}{a}$

The box is moving along the frictionless ramp in upward direction. Consider the sign convention according to the direction of forces.

$$\begin{gathered} -mg\sin \mathrm{θ}=ma \\ -g\sin \mathrm{θ}=a \\ \sin \mathrm{θ}=\frac{a}{-g} \\ \sin \mathrm{θ}=\frac{-2.50m/s^2}{-9.8m/s^2} \\ \mathrm{θ}=14.8^o \end{gathered}$$

To find normal acting on box, we can apply Newton’s second law along y axis as,

$$\begin{gathered} N-mg.\cos \mathrm{θ}=0 \\ N=mg.\cos \mathrm{θ} \\ =5.00kg×9.8m/s^2×\cos (14.8^o) \\ =47.4N \end{gathered}$$

Hence, the magnitude of the normal force on the box from the ramp is.