<u>A heavy object falls with the acceleration as a light object during free fall because:</u>
Heavy things have a large gravitational force and also have less acceleration. So both the effects exactly cancel and make the falling objects to have the same acceleration irrespective of mass.
Free fall is a unique motion which has only gravitational force that acts on an object. Objects that undergo free fall experience only have the influence of gravity and not any other force.
So when we apply newton's second law of gravity which is:
where,
F is the force
m is the mass
a is the acceleration
For example: When a 1000 kg elephant and a 1 kg rat fall from the same height,
The acceleration can be calculated as follows:
For elephant: F = 10000 N and m = 1000 kg. So,
For rat : F = 10 N and m = 1 kg
Thus, 
Hence, it shows that both the animals have the same acceleration irrespective of their mass.
Answer:
- The speed will be
Explanation:
We can use the following kinematics equation
where 
is the final speed, 
its the initial speed, a is the acceleration, and d the distance.
The force will be tripled, the force is:
in 1D
Now, for the original problem, we have
For the second problem, we have
starting from the rest, we have the same initial velocity.
As the force is tripled, we have:
But the mass its the same, so
So the acceleration its also tripled.
As the distance traveled by the arrow must also be the same, we have:
And this will be the speed from the arrow leaving the bow.
Answer:
70 N
21°
1.1 m/s²
Explanation:
Draw a free body diagram of the block. There are three forces:
Weight pulling straight down
Normal force pushing perpendicular to the incline
Friction force pushing parallel to the incline
Part 1
Sum the forces in the perpendicular direction:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
The block is at rest, so F = N μs:
F = N μs
F = mg μs cos θ
F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)
F = 70 N
Part 2
Sum the forces in the parallel direction (down the incline is positive):
∑F = ma
mg sin θ − F = 0
mg sin θ = N μs
mg sin θ = mg μs cos θ
tan θ = μs
θ = atan μs
θ = atan 0.38
θ = 21°
Part 3
Sum the forces in the parallel direction (this time, acceleration is not 0).
∑F = ma
mg sin θ − F = ma
mg sin θ − N μk = ma
mg sin θ − mg μk cos θ = ma
a = g (sin θ − μk cos θ)
a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)
a = 1.1 m/s²
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Answer: k = 2.07692
Explanation: Please find the attached files for the solution
