Question

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the horizontal. What is the magnitude of the acceleration of the crate?

Answer 1

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, $(\theta)$ = 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.

⇒ $F_x=Fcos(\theta)$          and             ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$                           ⇒ $F_y=15sin (15)$

⇒ Applying concept of  forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$        ...Newtons second law Fnet = ma

⇒ $a =\frac{F-f}{m}$              

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$     ...f is the friction which is zero here.

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.