A man pushes a 35.2 kg box across a frictionless floor with a force of 128 N. What is the acceleration of the box
2 answers:
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We can solve the problem by using the law of conservation of energy.
When the rocket starts its motion from the Earth surface, its mechanical energy is sum of kinetic energy and gravitational potential energy:
where
m is the rocket's mass
is the rocket initial speed
is the gravitational constant
is the Earth's mass
is the distance of the rocket from the Earth's center (so, it corresponds to the Earth's radius)
The mechanical energy of the rocket when it is very far from the Earth is just kinetic energy (because the gravitational potential at infinite distance from Earth is taken to be zero):
where
is the final speed of the rocket.
By equalizing the initial energy and the final energy, we can find the final velocity:
When the rocket starts its motion from the Earth surface, its mechanical energy is sum of kinetic energy and gravitational potential energy:
where
m is the rocket's mass
The mechanical energy of the rocket when it is very far from the Earth is just kinetic energy (because the gravitational potential at infinite distance from Earth is taken to be zero):
where
By equalizing the initial energy and the final energy, we can find the final velocity:
Answer:
x = 0.396 m
Explanation:
The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring
Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.
Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash
p₀ = m1 v₀₁
Moment after shock
= (m1 + m2)
p₀ =
m1 v₀₁ = (m1 + m2)
= v₀₁ m1 / (m1 + m2)
= 4.4 600 / (600 + 500)
= 2.4 m / s
With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring
Before compressing the spring
Em₀ = K = ½ (m1 + m2) ²
After compressing the spring
= Ke = ½ k x²
As there is no rubbing the energy is conserved
Em₀ =
½ (m1 + m2) ² = = ½ k x²
x = √ (k / (m1 + m2))
x = 2.4 √ (11/3000)
x = 0.396 m