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2 years ago

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A spring, with a spring constant of 4000 N/m, is oriented horizontally, and compressed by 10cm. When released, the spring launch

es a block of mass 1.0 kg along a 5.0-m horizontal section of track, where the coefficient of friction between the block and track is 0.20. The block then goes up a frictionless ramp angled at 60o with the horizontal. How high up the ramp does the block go before it starts to slide back down

1 answer:

cricket20 [7]2 years ago

5 0

Answer:

$d=1.2m$

Explanation:

From the question we are told that:

Spring constant $k= 4000 N/m$

Compressed $l_d= 10cm=>0.10$

Mass $m=1.0kg$

Length of horizontal section $l=5.0-m$

Coefficient of friction $\mu=0.20$

Angle $\theta=60 \textdegree$

Generally the equation for Kinetic Energy K.E is mathematically given by

$K.E=\mu mgL+mgdsin\theta$

$\frac{1}{2}k*l_d^2=\mu mgL+mgdsin\theta$

$\frac{1}{2}(4000)*0.1^2=0.2*1*9.8*5+1*9.8*d*sin60$

$d=1.2m$

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Where $\hat{v}$ is the unit vector of $\vec v$ , which is determined by the following expression:

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Where $\|\vec v\|$ is the norm of $\vec v$ , whose value can be found by Pythagorean Theorem.

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