Answer:
The bin moves 0.87 m before it stops.
Explanation:
If we analyze the situation and apply the law of conservation of energy to this case, we get:
Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)
F d = (0.5)(m)(Vi² - Vf²)
where,
F = Frictional Force = μR
but, R = Normal Reaction = Weight of Bin = mg
Therefore, F = μmg
Hence, the equation becomes:
μmg d = (0.5)(m)(Vi² - Vf²)
μg d = (0.5)(Vi² - Vf²)
d = (0.5)(Vi² - Vf²)/μg
where,
Vf = Final Velocity = 0 m/s (Since, bin finally stops)
Vi = Initial Velocity = 1.6 m/s
μ = coefficient of kinetic friction = 0.15
g = 9.8 m/s²
d = distance moved by bin before coming to stop = ?
Therefore,
d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)
<u>d = 0.87 m</u>
Answer: 4.08kg/J
Explanation: Please find the attached file for the solution
Answer:
1,211.1 kg.
Explanation:
the force of gravity is less on the moon than on earth, so if the man can lift 200kg on earth, he could lift a greater amount on the moon because there is less resistance from gravity.
To know the amount of mass he can lift on the moon, we first need to know the amount of weight that is equivalent to those 200kg here on earth. This because the weight of the object is equal to the force that must be applied to lift it, and that force is applied by the man and it will be the same here and on the moon.
We calculate weight using the formula:
where is the weight of the object (the force with which the earth attracts the object) is the mass and g the acceleration of gravity.
so
for earth the acceleration due to gravity is:
thus:
now we use this value to calculate the mass he can lift on the moon, since for the moon .
we use the same equation, w =mg substituting w = 1962N and :
he can lift 1,211.1 kg.
You can also find the result using the approximate value of the acceleration of gravity on the moon as g/6, where g is the acceleration on earth.
968 j or kg because it cannot lose any of it energy and cannot gain any extra mass from nothing
Answer:
The force applied 275 N in a direction parallel to the hill
Explanation:
Newton's second law is adequate to work this problem, in the annex we can see a free body diagram, where the weight (W) is vertical, the friction force (fr) is parallel to the surface and the normal (N ) is perpendicular to it. In general for these problems a reference system is taken that is parallel to the surface and the Y axis is perpendicular to it.
Let us decompose the weight into its two components, the angle T is taken from the axis and
Wx = W sin θ
Wy = W cos T
We write Newton's second law
∑ F = m a
X axis
The cyclist falls at a constant speed, which implies that the acceleration is zero
fr - W sin θ = 0
fr = mg sin θ
fr = 96 9.8 without 17
fr = 275 N
When the cyclist returns to climb the hill, he must apply the same force he has to overcome the friction force that always opposes the movement
. The force applied 275 N in a direction parallel to the hill