The final answer is already given in dark letter at the end of the question! 4 2

Question

The final answer is already given in dark letter at the end of the question! 4 2.5 kg k-250 N/m = 0.4 d = 0.50 m A 2.5 kg box is held released from rest 1.5 meters above the ground and slides down a frictionless ramp. It slides across a floor that is frictionless, except for a small section 0.5 meters wide that has a coefficient of kinetic friction of 0.4. At the left end, is a spring with spring constant 250 N/m. The box compresses the spring, and is accelerated back to the right. What is the speed of the box at the bottom of the ramp? v 5.42 m/s What is the maximum distance the spring is compressed by the box?x-0.505 m What is the maximum height to which the box returns on the ramp? hr 1.1 m

Explanation / Answer

using law of conservation of energy

energy at the top = energy at the bottom

m*g*h = 0.5*m*v^2

m cancels

g*h = 0.5*v^2

9.8*1.5 = 0.5*v^2

v = 5.42 m/s is the speed of the box at the bottom of the ramp

Now using work energy theorem

work done by the frictional force = change in kinetic energy

-mu_k*m*g*d = 0.5*m*(Vf^2-V^2)

m cancels

-0.4*9.8*0.5 = 0.5*(Vf^2-5.42^2)

vf = 5.04 m/s is the speed of the box to the right of the rough surface

Now using law of conservation of energy

Kinetic energy of the box = energy stored in spring

0.5*m*vf^2 = 0.5*k*x^2

0.5*2.5*5.04^2 = 0.5*250*x^2

x =0.505 m

while returning back the speed of the box after rough surface is V1

-mu_k*m*g*d = 0.5*m*(V1^2-Vf^2)

-0.4*2.5*9.8*0.5 = 0.5*2.5*(V1^2-5.04^2)

V1 = 4.63 m/s

Now using law of conservation of energy

energy at the bottom of the ramp = energy at the top of the ramp

0.5*m*v1^2 = m*g*h1

0.5*2.5*4.63^2 = 2.5*9.8*h1

h1= 1.1 m

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