A box is pushed against a spring (w/spring constant K) at the bottom of a fricti

Question

A box is pushed against a spring (w/spring constant K) at the bottom of a frictionless incline (angle from the horizontal theta). The spring is compressed at a distance x. After it is released, the box slides a distance up the incline and eventually comes to rest. Find an expression for the distance d that the box travels up the incline.If the mass of e box is 4kg, the spring constant is 900N/m, the incline angle is 5 degrees and the initial compression of the spring is 3cm, what is the distance the box travels up the incline?

Explanation / Answer

we have conservation of energy

0.5Kx^2 = mgh

so mgh = 0.5kx^2

h=kx^2 / 2mg

now we have distance d= h/sin theta = kx^2 / (2mg * sin theta)

second part

put the values to get the result

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