5-kg block is pushed down an inclined ramp with an initial velocity, vo. The blo

Question

5-kg block is pushed down an inclined ramp with an initial velocity, vo. The block slues own the ramp, through a distance d 0.4 m, before colliding with a spring of force constant, k 500 N/m. The ramp is inclined at an angle of 30° above the horizontal. On hitting the spring, the block compresses the spring through a maximum distance of 35 -em before momentarily coming to rest. The coefficient of kinetic friction between the block and the slope is 0.25 Use energy methods to calculate this initial velocity vo with which the block was pushed. 30°

Explanation / Answer

Given

mass m = 5 kg , d = 0.4 m , theta = 30 degrees , k = 500 N/m , mue_k = 0.25

the force acting on the block is  

(mg sin theta - mue_k*mg cos theta )d = 0.5*m(v^2-v0^2)

work energy theorem ==> work done = change in kinetic energy

(mg sin theta - mue_k*mg cos theta )d = 0.5*k*x^2 = 0.5*m(v^2-v0^2)

0.5*k*x^2 = 0.5*m(v^2-v0^2)

from conservation of energy

kinetic energy converts in to elastic potential energy, so that the blovk come to rest when the complete energy converted in to elastic potential energy

0.5*500*0.35^2 = 0.5*5(0-v0^2)

v0 = 3.5 m /s

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