Magnitude of normal force acting on the block is 7 N
Explanation:
10N = 1.02kg
Mass of the block = m = 1.02 kg
Angle of incline Θ
= 30°
Normal force acting on the block = N
From the free body diagram,
N = mgCos Θ
N = (1.02)(9.81)Cos(30)
N = 8.66 N
Rounding off to nearest whole number,
N = 7 N
Magnitude of normal force acting on the block = 7 N
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
A warmer climate is produced because of the gulf stream which begins at the tip of Florida and flows up all the way to NW Europe.
F = ma
Acceleration in this case is acceleration due to gravity so
a = 9.8 m/s^2
and mass = 0.57kg
So..
F = (0.57)(9.8)
F = 5.586 or 5.56 N
