Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
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.
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⇒ 
⇒
<em> ...Newtons second law Fnet = ma</em>
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⇒ Plugging the values.
⇒
<em>...f is the friction which is zero here.</em>
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⇒ 
Magnitude of the acceleration of the crate is 0.362 m/s^2.
For the scale od the paraboliczne curviczne Ur