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>
The number of protons in an atom is the elements atomic number.
Answer:
Ek= 35.392 kJ or 35.392 10³ J
Explanation:
Formula for Kinetic energy:
Ek= 
× m × 
since m= 65kg and v= 33m/s
Ek= × 65×
Ek= 35.392 kJ or 35.392 10³ J
Answer:
1. 0 J
2. 7500 J
3. 7500 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Final velocity (v₂) of car = 5 m/s
Original kinetic energy (KE₁) =?
Final kinetic energy (KE₂) =?
Work used =?
1. Determination of the original kinetic energy.
Mass (m) of car = 600 Kg
Initial velocity (v₁) of car = 0 m/s
Original kinetic energy (KE₁) =?
KE₁ = ½mv₁²
KE₁ = ½ × 600 × 0²
KE₁ = 0 J
Thus, the original kinetic energy of the car is 0 J.
2. Determination of the final kinetic energy.
Mass (m) of car = 600 Kg
Final velocity (v₂) of car = 5 m/s
Final kinetic energy (KE₂) =?
KE₂ = ½mv₂²
KE₂ = ½ × 600 × 5²
KE₂ = 300 × 25
KE₂ = 7500 J
Thus, the final kinetic energy of the car is 7500 J
3. Determination of the work used.
Original kinetic energy (KE₁) = 0
Final kinetic energy (KE₂) = 7500 J
Work used =?
Work used = KE₂ – KE₁
Work used = 7500 – 0
Work used = 7500 J
