Respuesta :
Answer:
The final speed of the block is 2.97 m/s
Solution:
As per the question:
mass of the ice block, m = 2 kg
distance, d = 0.750 m
Angle, [tex]\theta = 36.9^{\circ}[/tex]
Now,
We know that the work done can be given as the change in Kinetic energy of the block:
[tex]W = KE_{f} - KE_{i}[/tex]
Since, the bock slides we have velocity in the downward direction only along the slope:
[tex]W = \frac{1}{2}mv^{2} - 0 = \frac{1}{2}mv^{2}[/tex]
Now, the normal force is perpendicular to the displacement of the block, thus does not contribute to the work done, the only component of force that provides the work done is along the plane in the direction the block slides, thus:
[tex]F = mg\times sin\theta = mg\times sin36.9^{\circ}[/tex]
[tex]Work done,\ W = Fd = mgd\times sin36.9^{\circ}[/tex]
[tex]W = 2\times 9.8\times 0.750\times sin36.9^{\circ} = 8.826 J[/tex]
Now, by work energy theorem:
Work done, W = Kinetic energy, KE
Thus
[tex]KE = \frac{1}{2}mv^{2} = W[/tex]
[tex]v = \sqrt{\frac{2W}{m}}[/tex]
[tex]v = \sqrt{\frac{2\times 8.826}{2}} = 2.97 m/s[/tex]
The final speed of this block of ice is 2.972 m/s.
Given the following data:
- Mass of ice = 2 kg.
- Height = 0.750 m.
- Angle of inclination = 36.9°
To calculate the final speed of this block of ice, we would apply the work-energy theorem:
How to calculate the work done.
Mathematically, the work done on the block of ice is given by this formula:
[tex]W=Fd=mgdsin \theta\\\\W = 2 \times 9.8 \times 0.750 \times sin36.9\\\\W=14.7 \times 0.6004[/tex]
W = 8.83 Joules.
From the work-energy theorem, final speed is given by this formula:
[tex]V=\sqrt{\frac{2W}{m} } \\\\V=\sqrt{\frac{2 \times 8.83}{2} } \\\\V=\sqrt{8.83}[/tex]
V = 2.972 m/s.
Read more on work here: https://brainly.com/question/22599382
