A worker drives a spike into a rail tie with a 2.00-kg sledgehammer. This caused the internal energy of the hammer and spike to
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Answer:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
...(I)
Using gravitational potential energy
....(II)
Using energy of conservation
Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation
Hence, The maximum height above the point of release is 11.653 m.
An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let <em>θ</em> denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,
• the normal force with magnitude <em>n</em>
• the car's weight with magnitude <em>w</em>
and the net force points toward the center of the circle made by the turn, with centripetal acceleration
<em>a</em> = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²
Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,
∑ <em>F</em> (⟂) = <em>N</em> + <em>W</em> (⟂) = <em>m</em> <em>a</em> (⟂)
and
∑ <em>F</em> (//) = <em>W</em> (//) = <em>m a</em> (//)
Let the direction of <em>N</em> be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle <em>θ</em> with the banked curve, and <em>W</em> makes the same angle with the negative perpendicular axis, so that the equations above reduce to
<em>N</em> - <em>m g</em> cos(<em>θ</em>) = <em>m</em> <em>a</em> sin(<em>θ</em>)
and
<em>m g</em> sin(<em>θ</em>) = <em>m a</em> cos(<em>θ</em>)
The second equation is all we need at this point to find the ideal <em>θ</em>. The mass <em>m</em> cancels out, and we can solve for <em>θ</em> to get
tan(<em>θ</em>) = <em>a</em>/<em>g</em> ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615
→ <em>θ</em> ≈ 3.52°
