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Stells [14]
3 years ago
9

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
Physics
1 answer:
cricket20 [7]3 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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Answer:

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Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

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3 years ago
Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr
slava [35]

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Work done is equal to change in kinetic energy

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F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}

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