A ball that rolls on the ground is initially propelled with a speed of 45 km / h and after 10 seconds it stops. Assuming you los
1 answer:
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Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
How energy is conserved
Em₀ =
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)
Answer:
108.7 V
Explanation:
Two forces are acting on the particle:
- The external force, whose work is
- The force of the electric field, whose work is equal to the change in electric potential energy of the charge:
where
q is the charge
is the potential difference
The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:
and since the charge starts from rest, , so the formula becomes
In this problem, we have
is the work done by the external force
is the charge
is the final kinetic energy
Solving the formula for , we find