What is a change in the speed of an object called?
2 answers:
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You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.
There are at least 2 forces acting on the sled:
• its weight, pointing downward with magnitude <em>W</em> = <em>m g</em>
• the normal force, pointing perpendicular to the hill and away from the ground with mag. <em>N</em>
The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector <em>F</em> pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. <em>F</em>.
Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:
∑ (//) = <em>W</em> (//) = <em>m</em> <em>a</em> (//)
∑ (⟂) = <em>W</em> (⟂) + <em>N</em> = <em>m </em><em>a</em> (⟂)
where, for instance, <em>W</em> (//) denotes the component of the sled's weight in the direction parallel to the hill, while <em>a</em> (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -<em>F</em> to the first equation.
If the hill makes an angle of <em>θ</em> with flat ground, then <em>W</em> makes the same angle with the hill so that
<em>W</em> (//) = -<em>m g </em>sin(<em>θ</em>)
<em>W</em> (⟂) = -<em>m g</em> cos(<em>θ</em>)
So we have
<em>-m g </em>sin(<em>θ</em>) = <em>m</em> <em>a</em> (//) → <em>a</em> (//) = -<em>g </em>sin(<em>θ</em>)
<em>-m g</em> cos(<em>θ</em>) + <em>N</em> = <em>m </em><em>a</em> (⟂) → <em>a</em> (⟂) = 0
where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.
Then the acceleration vector is
<em>a</em> = <em>a</em> (//)
with magnitude
||<em>a</em>|| = <em>a</em> = <em>g </em>sin(<em>θ</em>).
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