The cable of the 1300 kg elevator cab in the figure snaps when the cab is at res
ID: 2259211 • Letter: T
Question
The cable of the 1300 kg elevator cab in the figure snaps when the cab is at rest at the first floor, where the cab bottom is a distance d = 3.3 m above a spring of spring constant k = 0.33 MN/m. A safety device clamps the cab against guide rails so that a constant frictional force of 2.9 kN opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance x that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance (above the point of maximum compression) that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)
Explanation / Answer
Given Mass of the cab, m = 1300 kg The cab bottom is at a distance above the spring is, d = 3.3 m Spring constant, k = 0.33 *10^6 N/m Frictional force, f = 2.9 *10^3 N a)By law of conservation of energy, mgd = 1/2 mv^2 + f * d (1300 kg) (9.8 m/s^2) (3.3 m) = 1/2 (1300 kg) v^2 + (2.9 *10^3 N) (3.3 m) v = 7.06 m/s
is the speed of the cab just before it hits the spring b)
Again applying law of conservation of energy 1/2 mv^2+ mgx = 1/2 kx^2 + f * x 1/2 (1300 kg) (7 m/s)^2 +(1300 kg) (9.8 m/s^2) x = 1/2 ( 0.33 *10^6 N/m) x^2 + (2.9 *10^3 N) x 165000x^2 - 9840 x - 31850 = 0 Solving this equation, we get x = 0.47 m c)
Let it bounce to a height h , then applying law of conservation of energy 1/2 kx^2 = mgh + f *h 1/2 ( 0.33 *10^6 N/m) (0.47 m)^2 = (1300 kg) (9.8 m/s^2) h + (2.9 *10^3 N) h h = 2.33 m d)
Let D be the distance covered by the cab after bouncing back up the shaft. D can be calculated by using Newton's second law mg - kD = 0 D = (1300 kg) (9.8 m/s^2) / ( 0.33 *10^6 N/m) D = 0.0386 m Total distance travelled by the cab before coming o rest is, 1/2 kD^2 + f *Dtotal = m g (D + d) 1/2 ( 0.33 *10^6 N/m) (0.0386m)^2 +(2.9 *10^3 N) *Dtotal = (1300 kg) (9.8 m/s^2)
Dtotal = 2.07 m
(0.0386+2.07 m) Dtotal = 2.1 m Given Mass of the cab, m = 1300 kg The cab bottom is at a distance above the spring is, d = 3.3 m Spring constant, k = 0.33 *10^6 N/m Frictional force, f = 2.9 *10^3 N a)
By law of conservation of energy, mgd = 1/2 mv^2 + f * d (1300 kg) (9.8 m/s^2) (3.3 m) = 1/2 (1300 kg) v^2 + (2.9 *10^3 N) (3.3 m) v = 7.06 m/s
is the speed of the cab just before it hits the spring b)
Again applying law of conservation of energy 1/2 mv^2+ mgx = 1/2 kx^2 + f * x 1/2 (1300 kg) (7 m/s)^2 +(1300 kg) (9.8 m/s^2) x = 1/2 ( 0.33 *10^6 N/m) x^2 + (2.9 *10^3 N) x 165000x^2 - 9840 x - 31850 = 0 Solving this equation, we get x = 0.47 m c)
Let it bounce to a height h , then applying law of conservation of energy 1/2 kx^2 = mgh + f *h 1/2 ( 0.33 *10^6 N/m) (0.47 m)^2 = (1300 kg) (9.8 m/s^2) h + (2.9 *10^3 N) h h = 2.33 m d)
Let D be the distance covered by the cab after bouncing back up the shaft. D can be calculated by using Newton's second law mg - kD = 0 D = (1300 kg) (9.8 m/s^2) / ( 0.33 *10^6 N/m) D = 0.0386 m Total distance travelled by the cab before coming o rest is, 1/2 kD^2 + f *Dtotal = m g (D + d) 1/2 ( 0.33 *10^6 N/m) (0.0386m)^2 +(2.9 *10^3 N) *Dtotal = (1300 kg) (9.8 m/s^2)
Dtotal = 2.07 m
(0.0386+2.07 m) Dtotal = 2.1 m c)
Let it bounce to a height h , then applying law of conservation of energy 1/2 kx^2 = mgh + f *h 1/2 ( 0.33 *10^6 N/m) (0.47 m)^2 = (1300 kg) (9.8 m/s^2) h + (2.9 *10^3 N) h h = 2.33 m d)
Let D be the distance covered by the cab after bouncing back up the shaft. D can be calculated by using Newton's second law mg - kD = 0 D = (1300 kg) (9.8 m/s^2) / ( 0.33 *10^6 N/m) D = 0.0386 m Total distance travelled by the cab before coming o rest is, 1/2 kD^2 + f *Dtotal = m g (D + d) 1/2 ( 0.33 *10^6 N/m) (0.0386m)^2 +(2.9 *10^3 N) *Dtotal = (1300 kg) (9.8 m/s^2)
Dtotal = 2.07 m
(0.0386+2.07 m) Dtotal = 2.1 m d)
Let D be the distance covered by the cab after bouncing back up the shaft. D can be calculated by using Newton's second law mg - kD = 0 D = (1300 kg) (9.8 m/s^2) / ( 0.33 *10^6 N/m) D = 0.0386 m Total distance travelled by the cab before coming o rest is, 1/2 kD^2 + f *Dtotal = m g (D + d) 1/2 ( 0.33 *10^6 N/m) (0.0386m)^2 +(2.9 *10^3 N) *Dtotal = (1300 kg) (9.8 m/s^2)
Dtotal = 2.07 m
(0.0386+2.07 m) Dtotal = 2.1 m
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