Part 4 A 1m by 1m block, M = 5.00 kg, is 25.0 m up an incline from a spring, who

Question

Part 4
A 1m by 1m block, M = 5.00 kg, is 25.0 m up an incline from a spring, whose spring constant is k = 100. N/m. Make the length of the spring to be 3.00 m (or sufficiently long enough) for its equilibrium length. The object is released from rest. The incline makes an angle of 30.0° with respect to the horizontal and the static and kinetic coefficients of friction are 0.400. Determine the amount that the spring is compressed. Simulate on IP and determine the percent error.

Explanation / Answer

Let the height at which the block comes to rest be taken as the 0 potential point.

Let it gets compressed by d length

So, the energy stored by the spring , PEs = 0.5*k*d^2

where k = 100 N/m

Now, the gravitational potential energy at the initial posistion ,

PEg = mg*(25 +d)*sin(30)

Work done by friction:

Wf = umg*cos(30 deg)*(25+d)

By conservation of energy

PEs + Wf = PEg

So, 0.5*k*d^2 + 0.4*5*9.8*cos(30 deg)*(d+25) = mg*(25 +d)*sin(30)

So, 0.5*100*d^2 + 0.4*5*9.8*cos(30 deg)*(d+25)= 5*9.8*(25+d)*sin(30 deg)

So, d = 2.02 m <-------answer

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