Jake and Amir are both standing on identical skateboards and
1 answer:
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To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:
Here,
m = mass
v = Velocity
Our values are given as,
Replacing,
Therefore the mechanical energy lost due to friction acting on the runner is 907J
Explanation:
First consider that each hand works as a fulcrum: a pivot point where the barbell can rotate.
Now consider only the left hand. If the center of mass of the barbell is between hands (in the middle) it is displaced respect the fulcrum, therefore the weight which is pushing the bar downwards becomes a rotational force. The same thing happens to the other hand. Now, if more weight is added to the left hand the center of mass is displaced towards the left hand and depending how much weight is added, the center of mass will change its position and therefore the torque each hand experiences changes.
If the center of mass is still between hands: The torque remains almost the same changing only the magnitudes but not the direction.
If the center of mass is on the hand: there is no torque for the left hand because there is no leaver.
If the center of mass is to the left: now the torque changes direction and both hands need to stop it in the same direction.
(see diagram below)
Answer:
v₀ = 292.3 m / s
Explanation:
Let's analyze the situation, on the one hand we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy, as the data they give us are Let's start with this second part.
We write the mechanical energy when the shock has passed the bodies
Em₀ = K = ½ (m + M) v²
We write the mechanical energy when the spring is in maximum compression
=
= ½ k x²
Em₀ =
½ (m + M) v² = ½ k x²
Let's calculate the system speed
v = √ [k x² / (m + M)]
v = √[154 0.83² / (0.012 +0.104) ]
v = 30.24 m / s
This is the speed of the bullet + Block system
Now let's use the moment to solve the shock
Before the crash
p₀ = m v₀
After the crash
= (m + M) v
The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved
p₀ =
m v₀ = (m + M) v
v₀ = v (m + M) / m
let's calculate
v₀ = 30.24 (0.012 +0.104) /0.012
v₀ = 292.3 m / s