A 2.00 kg block slides down and inclined plane that is 3.25 m long and is angled

Question

A 2.00 kg block slides down and inclined plane that is 3.25 m long and is angled 25.0 degrees above the horizontal with an initial speed of 1.34 m/s. At the bottom of the inclined plane is a flat smooth surface with a spring at the end. The spring constant of the spring is 200 N/m, and the coefficient of friction between the inclined plane and the block is 0.231.

Find the speed of the block at the botton of the inclined plane.

I need help with the concepts in this problem, any explanation of the equations and concepts you used would be very helpful.

Explanation / Answer

using conservation of energy,
the initial K.E+P.E will be spent on the work done by frictional force and the final K.E
so,
.5*2*(1.34)^2 + 2*9.8*3.25*sin25= .5*2*v^2 + .231*2*9.8*cos25*3.25
so, v= 3.92 m/s    <-----answer

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