Answer:
Acceleration of the crate is 0.362 m/s^2.
Explanation:
Given:
Mass of the box, m = 40 kg
Applied force, F = 15 N
Angle at which the force is applied, [tex](\theta)[/tex] = 15°
We have to find the magnitude of the acceleration.
Let the acceleration be "a".
FBD is attached with where we can see the horizontal and vertical component of force.
⇒ [tex]F_x=Fcos(\theta)[/tex] and ⇒ [tex]F_y=Fsin (\theta)[/tex]
⇒ [tex]F_x=15cos(15)[/tex] ⇒ [tex]F_y=15sin (15)[/tex]
⇒ Applying concept of forces.
⇒ [tex]\sum F_x=F_n_e_t =F-f[/tex]
⇒ [tex]F_n_e_t =F-f[/tex]
⇒ [tex]ma =F-f[/tex] ...Newtons second law Fnet = ma
⇒ [tex]a =\frac{F-f}{m}[/tex]
⇒ Plugging the values.
⇒ [tex]a =\frac{15cos(15)-0}{40}[/tex] ...f is the friction which is zero here.
⇒ [tex]a =\frac{14.48}{40}[/tex]
⇒ [tex]a=0.362\ ms^-^2[/tex]
Magnitude of the acceleration of the crate is 0.362 m/s^2.
