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CT8-7. A block initially at rest is allowed to slide down a frictionless ramp and attains a
speed v at the bottom. To achieve a speed 2v at the bottom, how many times as high
must the new ramp be?
Pink: 2
Yellow: 4
 14
Green: 2
Purple: none of these.
Blue: 3
Answer: 4 times as high. By conservation of energy, PEi = KEf, mgh = (1/2) mv2.
So, h is proportional to v2. If v doubles, v2 quadruples and so does h.
CT8-8. A mass slides down a frictionless ramp of height h and hits a carpet with kinetic
friction coefficient K = 1.0. Its initial speed is zero. How far does the mass slide along
the carpet?
Pink: h
Yellow: Less than h
Purple: more than h
Answer: h. Use conservation of energy: Initial energy at top of hill = mgh = energy at
bottom of hill (before puck hits rug) = (1/2)m2 = heat generated by friction as puck slows
and stops = magnitude of work done by friction = Wfric  K N x  K mg x . So we have
the equation mgh = Kmgx or h = Kx = x (since K=1).
CT8-9. An object moves along the x-axis . The potential energy U(x) vs. position x is
shown below.
When the object is at postion x=A, which of the following statements must be true.
Pink: The velocity is negative, i.e. along the -x direction.
Yellow: The acceleration is negative.
Green: The total energy is negative.
Blue: The kinetic energy is positive.
Purple: None of these statements is always true.
Answer: The acceleration is negative. In 1D, F = - dU/dx. The slope of the curve dU/dx
is positive at x=A, so F = ma = -dU/dx is negative there. The force F and the
acceleration a = F/m are negative at x=A (meaning their directions are to the left, toward
negative x.)
The kinetic energy might be zero at x=A, since the total energy of the system was never
specified. That is, the total energy might be such that x=A is a turning point.