A compressed spring, with spring constant 1900 N/m, is sitting on a level, froze

Question

A compressed spring, with spring constant 1900 N/m, is sitting on a level, frozen lake. One end of the spring is anchored firmly at one end to a large rock that pokes through the ice, and the other touches a 3.5 kg, rectangular block of ice. The spring is compressed a distance of 15 cm. the ice is very smooth and is slightly wet, making it, for all practical purposes, frictionless. a) what is the potential energy of the spring when it is compressed? b) the spring is released, which then pushes the block across the ice. What is the kinetic energy of the ice block as it moved across the ice? c) what is the speed of the block of ice, once it is moving on the frozen lake? d) the block of ice now slides up an icy ramp (also smooth and slightly wet). What is the gravitational potential energy of the block when it reaches the highest point on the ramp? e) when the block is at its highest point on the ramp, how high above the frozen lake will it be? A compressed spring, with spring constant 1900 N/m, is sitting on a level, frozen lake. One end of the spring is anchored firmly at one end to a large rock that pokes through the ice, and the other touches a 3.5 kg, rectangular block of ice. The spring is compressed a distance of 15 cm. the ice is very smooth and is slightly wet, making it, for all practical purposes, frictionless. a) what is the potential energy of the spring when it is compressed? b) the spring is released, which then pushes the block across the ice. What is the kinetic energy of the ice block as it moved across the ice? c) what is the speed of the block of ice, once it is moving on the frozen lake? d) the block of ice now slides up an icy ramp (also smooth and slightly wet). What is the gravitational potential energy of the block when it reaches the highest point on the ramp? e) when the block is at its highest point on the ramp, how high above the frozen lake will it be?

Explanation / Answer

a) PE = 0.5*k*x^2 = 0.5*1900*0.15^2 = 21.375 J

B) using conservation of energy

KE = 21.375 J

C)0.5*m*v^2 = 21.375

v = sqrt(21.375/(0.5*m)) = sqrt(21.375/(0.5*3.5)) = 3.5 m/s

D) using conservation of energy

gravitational PE = 21.375 J

E) m*g*h = 21.375

h = 21.375/(m*g) = 21.375/(3.5*9.81) = 0.622 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.