A small block of mass 0.8 kg is launched by a compressed spring with force constant k=800 N/m. The initial compression of the spring is 0.22 m. The block slides along a horizontal frictionless surface and then up an inclined plane that makes an angle θ=40 degrees with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is μk=0.1. Use g=9.8m/s2.

Find the maximum vertical height h reached by the block.

Question

A small block of mass 0.8 kg is launched by a compressed spring with force constant k=800 N/m. The initial compression of the spring is 0.22 m. The block slides along a horizontal frictionless surface and then up an inclined plane that makes an angle θ=40 degrees with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is μk=0.1. Use g=9.8m/s2.

Find the maximum vertical height h reached by the block.

jaredstone4 · Answer

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anonymous · Answer

When the block is launched by the spring, the elastic potential energy turns into kinetic energy. 1/2*(800)(0,22^2) = 1/2*(0,8)(V^2) ; V=6,957m/s
The horizontal frictionless surface doesn't absorb any energy...
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Now, on the inclined plane, it has 2 forces pushing against the direction of the speed: The 0,8*9,8*sin(40) N and the friction force, which is 0,1*0,8*9,8*cos(40)N. Both of the forces add up to 5,64. This translates to an acceleration of 7,05m/s^2 opposite to the movement.
Using the linear movement equations: Vf^2 = Vi^2 + 2a*d
0=(6,957)^2 +2(-7,05)*d, giving a distance of 3,43m.
The h you need is just d*sin(40)=2,2 aprox.

jaredstone4 · Answer

That makes a lot of sense. Thanks so much!

anonymous · Answer

np :D