Answer:

Explanation:
For this case we can use the second law of Newton given by:

The friction force on this case is defined as :

Where N represent the normal force,
the kinetic friction coeffient and a the acceleration.
For this case we can assume that the only force is the friction force and we have:

Replacing the friction force we got:

We can cancel the mass and we have:

And now we can use the following kinematic formula in order to find the distance travelled:

Assuming the final velocity is 0 we can find the distance like this:

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,
= 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.
⇒
and ⇒ 
⇒
⇒ 
⇒ Applying concept of forces.
⇒
⇒ 
⇒
<em> ...Newtons second law Fnet = ma</em>
⇒
⇒ Plugging the values.
⇒
<em>...f is the friction which is zero here.</em>
⇒ 
⇒ 
Magnitude of the acceleration of the crate is 0.362 m/s^2.
I think the correct answer among the choices listed above is option B. The acceleration of gravity is a constant equal to 9.8 meters per second squared. You can see that all things fall at this rate if there is no air resistance in a system or when in a vacuum.