Answer:
[tex]a=-5 m/s^2[/tex]
Explanation:
We are given that Initial speed of block=[tex]v_0[/tex]
[tex]\theta=30^{\circ}[/tex]
We have to find the approximate acceleration of the block at the instant that it reaches highest point on the inclined plane.
By Newton's second law
[tex]F+mg sin\theta=0[/tex]
We know that [tex]F=ma[/tex]
[tex]ma=-mg sin\theta[/tex]
[tex]a=-gsin\theta[/tex]
g=[tex]9.8 m/s^2[/tex]
Substitute the values then we get
[tex]a=-9.8 sin30^{\circ}[/tex]
[tex]a=-9.8\times \frac{1}{2}=-4.9\approx -5m/s^2[/tex]
[tex] sin 30^{\circ}=\frac{1}{2}[/tex]
Hence, the approximate acceleration of the block=[tex]-5 m/s^2[/tex]
Where negative sign indicates that the block is moving in downward direction.
We have that for the Question "A block is projected up a frictionless plane with an initial speed vo. The plane is inclined 30° above the horizontal. What is the approximate acceleration of the block at the instant that it reaches its highest point on the inclined plane?" it can be said that the approximate acceleration of the block at the instant that it reaches its highest point on the inclined plane is
a=4.9m/s^2
From the question we are told
A block is projected up a frictionless plane with an initial speed vo. The plane is inclined 30° above the horizontal. What is the approximate acceleration of the block at the instant that it reaches its highest point on the inclined plane?
Generally the equation for the acceleration is mathematically given as
[tex]mgsin\theta=ma\\\\Therefore\\\\a=gsin30\\\\a=9.8/2\\\\[/tex]
a=4.9m/s^2
Therefore
the approximate acceleration of the block at the instant that it reaches its highest point on the inclined plane is
a=4.9m/s^2
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