Answer:
The lighter frog goes higher than the heavier frog.
The lighter frog is moving faster than the heavier frog
Explanation:
If both frogs have the same kinetic energy when they leave the ground, the following equality applies:
Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:
When h= hmax, the object comes momentarily to an stop, so vf =0
Solving for hmax:
As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.
As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.
Answer:
73.67 m
Explanation:
If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:
,
Where 
its our initial height, 
our initial speed, a the acceleration and t the time that has passed.
For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.
We can plug this values in our equations, to obtain:
note that the acceleration point downwards, hence the minus sign.
Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:
and obtain:
Plugin this time on our first equation we find:
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
or
where <em>K</em> is the block's kinetic energy at the equilibrium point,
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
Compute the work performed by friction:
By Newton's second law, the net vertical force on the block is
∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0 ==> <em>n</em> = <em>mg</em>
where <em>n</em> is the magnitude of the normal force from the surface pushing up on the block. Then if <em>f</em> is the magnitude of kinetic friction, we have <em>f</em> = <em>µmg</em>, where <em>µ</em> is the coefficient of kinetic friction.
So we have
Answer:
1. 
= 5.45 m/s
, 2. K = 326.73 J and 3. h = 152 cm
Explanation:
R1. Let's use the conservation of the moment, for this we define a system formed by the two bodies, the pill plus the hanging mass,
Where the mass of the tablet (m = 2 kg) and the hanging mass (M = 38 Kg)
Initial, before crash
po = m v₀₁ + 0
Final, just after the crash
= (m + M)
The moment is preserved
p₀ =
m v1o = (m + M)
= m / (m + M) v1o
= 2/(2+20) 60
= 5.45 m/s
R2 The kinetic energy is given, in our case, after the collision
K = ½ (m + M) ²
K = ½ (2 +20) 5.45²
K = 326.73 J
R3 Let's use the conservation of mechanical energy, after the crash. Let's look for energy at two points the lowest and the highest point
Lowest point
Em₀ = K = ½ (m + M) ²
Highest point
= U = mg h
Em₀ =
½ (m + M) ² = (m + M) g h
h =² / 2g
h = 5.45²/2 9.8
h = 1.52 m (100cm / 1m)
h = 152 cm
Given: Mass m = 60 Kg
Weight W = 96 N
Required: Acceleration due to gravity, g = ?
Formula: W = mg
g = W/g
g = 96 Kg.m/s²/60 Kg (note: this is the derive unit for Newton "N")
g = 1.6 m/s²
