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4 years ago

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A block of mass m1 = 290 g is at rest on a plane that makes an angle θ = 30° above the horizontal. The coefficient of kinetic fr

iction between the block and the plane is μk = 0.10. The block is attached to a second block of mass m2 = 200 g that hangs freely by a string that passes over a frictionless and massless pulley. Find its speed when the second block has fallen 30.0 cm.

1 answer:

GrogVix [38]4 years ago

5 0

Answer:

The speed of the block when is fallen 30cm

v=0.726 $\frac{m}{s}$

Explanation:

∑F= (m2)g - ƒ - (m1)g*sin(θ) = (m1)a

g = 9.81 m/s²

ƒ = μN = μ(m1)g

$(m2)*g - u*(m1)*g - (m1)*g*sin(\alpha ) = (m1)*a$

(0.200)(9.81) - (0.1)(0.290)(9.81) - (0.290)(9.81)sin(30°) = (0.290)a

$(0.200kg)(9.81\frac{m}{s^{2}}) - (0.1)(0.290kg)(9.81\frac{m}{s^{2}}) - (0.290kg)(9.81\frac{m}{s^{2}})sin(30°) = (0.290kg)*a$

$0.511101=0.29*a\\a=0.879\frac{m}{s^{2} }$

$v_{f}^{2}=v_{o}^{2}+2*a(x_{f}^{2}-v_{o}^{2})\\v_{o}=0\\v_{o}=0\\v_{f}^{2}=2*a(x_{f})\\v_{f}=\sqrt{2*a(x_{f})}\\v_{f}=\sqrt{2*0.879\frac{m}{s^{2}}*0.30m} \\v_{f}=0.726 \frac{m}{s}$

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