A small block of mass m = 0.50 kg is fired with an initial speed of v0 = 4.3 m/s
ID: 2217396 • Letter: A
Question
A small block of mass m = 0.50 kg is fired with an initial speed of v0 = 4.3 m/s along a horizontal section of frictionless track, as shown in the top portion of Figure P7.58. The mass then moves along the frictionless semicircular, vertical tracks of radius R = 1.4 m. Figure P7.58 (a) Determine the magnitude of the force exerted by the track on the block at points A and B. track force at A N track force at B N (b) The bottom of the track consists of a section (L = 0.37 m) with friction. Determine the coefficient of kinetic friction between the block and that portion of the bottom track if the block just makes it to point C on the first trip. [Hint: If the block just makes it to point C, the force of contact exerted by the track on the block at that point should be zero.]Explanation / Answer
If the block just makes it then, as the hint suggests, then all of the force keeping it in circular motion at the end is due to mg downward. The centripetal acceleration is g, with a circular path of radius R and an exit speed of v_f. The centripetal acceleration equation is then: g = v_f^2 / R .... or: v_f = sqrt(gR) For the problem to even be possible, this must be less than the initial velocity v_0 = 4.3 m/s, and the magnitude of the energy lost in the one trip is: ?K = m*v_0^2/2 - m*v_f^2/2 = (m/2)(v_0^2 - v_f^2) If the only energy loss was due to friction over a length L with kinetic coefficient of friction µ, then the energy loss must be the magnitude of the work done by friction: ?K = µmgL µ = (m/2)(v_0^2 - v_f^2)/(mgL) µ = (v_0^2 - gR)/(2gL) Sanity check: all terms in the numerator and denominator have dimensionality (length/time)^2, so the final result is dimensionless, as expected. Since v_0 > v_f, the result is positive, as expected. Time to plug numbers and calculate.