Answer:
F = 39.2 N
Explanation:
Since, the object is in uniform motion. Therefore, the frictional force on object will be:
Frictional Force = μk N = μk mg
where,
μk = coefficient of kinetic friction = 0.2
m = mass of crate = 10 kg
g = 9.8 m/s²
Therefore,
Frictional Force = (0.2)(10 kg)(9.8 m/s²)
Frictional Force = 19.6 N
The horizontal component of force must be equal to this frictional force to continue the uniform motion:
F Sin 30° = 19.6 N
F = 19.6 N/Sin 30°
<u>F = 39.2 N</u>
All forms of matter are comprised of atoms. Wind in itself is the movement of particles(air). Wind does not have particles since it is an action and not a form of matter.
(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.
(b) The acceleration of the child-wagon system is 0.33 m/s².
(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.
<h3>
Net force on the third child</h3>
Apply Newton's second law of motion;
∑F = ma
where;
- ∑F is net force
- m is mass of the third child
- a is acceleration of the third child
∑F = 96 N - 75 N - 12 N = 9 N
Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;
- the wagon
- the children outside the wagon
<h3>Free body diagram</h3>
→ → Ф ←
1st child friction wagon 2nd child
<h3>Acceleration of the child and wagon system</h3>
a = ∑F/m
a = 9 N / 27 kg
a = 0.33 m/s²
<h3>When the frictional force is 21 N</h3>
∑F = 96 N - 75 N - 21 N = 0 N
a = ∑F/m
a = 0/27 kg
a = 0 m/s²
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