A 10kg stone is moving at 15m/s at the top of a 20m high, frictionless hill. At

Question

A 10kg stone is moving at 15m/s at the top of a 20m high, frictionless hill. At the bottom of the hill, it encounters a rough, horizontal surface on which it travels for 100m before reaching a spring with a force constant of2N/m. The coefficients of static and kinetic friction between the stone and the horizontal ground are 0.8, and 0.2, respectively, How fast is the stone going when it reaches the bottom of the hill? How tar does w spring get compressed in bringing the stone to a stop? What is the force exerted by the spring on the Stone when it has been fully compressed? Is this enough to overcome the maximum value of the frictional force, and what is this value?

Explanation / Answer

Speed at the bottom of the hill is vb

using law of conservation of energy

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

0.5*10*15^2 + 10*9.81*20 = 0.5*10*vb^2

vb = 24.84 m/s

after that it travels frional surface

Work done by the frictional force = change in kinetic energy

-fk*S= 0.5*m*(vf^2-vb^2)

-mu_k*m*g*S = 0.5*10*(vf^2-24.84^2)

-0.2*10*9.81*100 = 0.5*10*(vf^2-24.84^2)

vf = 15 m/s is the speed before hitting spring

then again using law of conservation of energy

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

x = sqrt(m/k)*vf = sqrt(10/2)*15 = 33.54 m is the compression of the spring

C) F = k*x

F = 2*33.54 = 67.08 N

d) maximum value of frictional force = mu_s*m*g = 0.8*10*9.81 = 78.48 N

not enough to overcome the maximum frictiuonal force

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