Answer:
ugmd = 1/2 kx²
d = (1/2 kx²) / (ugm)
= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)
= 7.4 m
ugmd = 1/2 mv²
v = √2ugd
= √(2(0.23)(9.81 m/s²)(7.4 m)
= 5.8 m/s
Explanation:
Answer:
There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.
Answer:
D
friction acts in the opposite direction of motion but does not affect the motion of the object
We will apply the conservation of linear momentum to answer this question.
Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Note that v₁ and v₁' is the velocity of m₁ before and after the collision.
Let's choose m₁ and v₁ to represent the bullet's mass and velocity.
m₂ and v₂ represents the wood block's mass and velocity.
The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'
m₁v₁ + m₂v₂ = m₁v' + m₂v'
m₁v₁ + m₂v₂ = (m₁+m₂)v'
Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.
m₁v₁ = (m₁+m₂)v'
Here are the given values:
m₁ = 0.005kg
v₁ = 500m/s
m₂ = 5kg
Plug in the values and solve for v'
0.005×500 = (0.005+5)v'
v' = 0.4995m/s
v' ≅ 0.5m/s
