A 50 kg boy runs and jumps with a forward velocity of 1.5 m/s into a 125 kg stationary boat.
2 answers:
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1) The potential energy is the most at the highest position and the least at the equilibrium position
2) The kinetic energy is the most at the equilibrium position and the least at the highest position
Explanation:
1)
The potential energy of an object is the energy possessed by the object due to its position in a gravitational field; mathematically, it is given by
where
m is the mass of the object
g is the strength of the gravitational field
h is the height of the object relative to the ground
For the pendulum in this problem, m is the mass of the bob, and h is the height of the above relative to the ground. We see from the formula that the potential energy is directly proportional to the height:
This means that:
- The potential energy is the most when the bob is at the highest position
- The potential energy is the least when the bob is at the equilibrium position, which is the lowest position
2)
We can solve this part by applying the law of conservation of energy: in fact, the total mechanical energy of the pendulum (sum of potential and kinetic energy) is constant at any time during the motion,
where KE is the kinetic energy.
From the equation above, we observe that:
- When PE is maximum, KE must be at minimum
- When PE is minimum, KE must be maximum
Therefore, this implies that:
- The kinetic energy is the most when the potential energy is the least, i.e. at the equilibrium position
- The kinetic energy is the least when the potential energy is the most, i.e. at the highest position
Learn more about kinetic and potential energy:
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Answer:
a) time taken to catch up with speeding car is 12.25 secs
b) the police car will travel 273.8 m to catch up with the speeding car
Explanation:
Given that;
speed of car = 50 mi/hr = 22.352 m/s
acceleration of police car = 10 mi/hr = 4.47 m/s²
= 70 mi/hr = 31.29 m/s
Now time taken to reach maximum speed is t₁
so
=
+ at₁
we substitute
31.29 = 0 + 4.47t₁
t₁ = 31.29 / 4.47
t₁ = 7 sec
now
d₁ = 0 + 1/2 × at₁²
d₁ = 0 + 1/2 × 0 + 4.47×(7)²
d₁ = 109.5 m
so distance travelled by the speeding car in time t₁ will be
=
× t₁
we substitute
= 22.352 × 7
= 156.46 m
now distance between polive car and speeding car
Δd = - d₁
Δd = 156.46 - 109.5
Δd = 46.96 m
time taken to cover Δd will be
t₂ = Δd / ( -
)
t₂ = 46.96 / ( 31.29 - 22.352 )
t₂ = 46.96 / 8.938
t₂ = 5.25 sec
distance travelled by the police in time t₂ will be
d₂ = × t₂
d₂ = 31.29 × 5.25
d₂ = 164.3 m
a) How long will it take before the officer catches up to the speeding car;
time taken to catch up with speeding car;
t = t₁ + t₂
t = 7 + 5.25
t = 12.25 secs
Therefore, time taken to catch up with speeding car is 12.25 secs
b) how far will it have travelled in order to do so;
distance = d₁ + d₂
distance = 109.5 + 164.3
distance = 273.8 m
Therefore, the police car will travel 273.8 m to catch up with the speeding car