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A block of mass 3.5 kg slides down a frictionless inclined plane of length 6.4 m that makes an angle of 30 degrees with the horizontal. if the block is released from rest at the top of the incline, what is its speed at the bottom

Respuesta :

MaximusMeridius
mass = 3,5 kg
weight = 35 N
distance = 6,4 m

Weight on X axis: Wx = weight.sine30
Wx = 35.0,5 = 17,5 N

F = m.a
17,5 = 3,5.a
a = 17,5/3,5
a = 5 m/s²

V² = V0² + 2a.d
V² = 0 + 2.5.6,4
V = √64
V = 8 m/s

Answer:

The speed of the block at the bottom is 7.92 m/s.

Explanation:

Given that,

Mass of block = 3.5 kg

Length l = 6.4 m

Horizontal angle = 30°

According to figure,

Using balance equation

[tex]ma=mg\ sin\theta[/tex]

Here, m = mass of block

a = acceleration of the block

[tex]a = g\ sin\theta[/tex]

Using equation of motion

[tex]v^2=u^2+2as[/tex]

[tex]v^2=0+2\times g\ sin\theta\times6.4[/tex]

[tex]v^2= 2\times g\ sin\ 30^{\circ}\times 6.4[/tex]

[tex]v^2=2\times9.8\times0.5\times6.4[/tex]

[tex]v^2=62.72[/tex]

[tex]v=7.92\ m/s[/tex]

Hence, The speed of the block at the bottom is 7.92 m/s.

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