2018 Spring Semester PHYS2321 -02 Test 4 Student\'s T # Student\'s Name_ 1, The
ID: 2038702 • Letter: 2
Question
2018 Spring Semester PHYS2321 -02 Test 4 Student's T # Student's Name_ 1, The cable of an elevator of mass m 1000 kg (shown in the Figure on th- right) brock and is falling down with a velocity of u,- 6.00 m/s before touching the spring. It is stopped by a frictional force F 20,000 N between the wall and the elevator and a spring of spring constant k after compressed the spring for 2.00 m. If y 0 at point 1, Find (a) The work done by the frictional force. (b) The total energy of the elevator before it touches the spring. (c) The total energy of the spring after it compresses the spring for 2.00 m (d) Use the work and energy relation to find the spring constant k (10 points each for (a), (b), and (c), 20 points for (d)Explanation / Answer
mass of the elevator (m) = 1000 Kg.
Velocity of the elevator = 6.00 m/s
The elevator is a freely falling body with a velocity of 6.00 m/s.
hence v2 = 2 * 9.8 * L = > L = 36 / 19.6 = 0.094
Drop height (L) = 9.4 cms
Frictional Force acting on the spring = 20000 N
Length of the compressed spring = 2.00 m
The total kinetic energy the spring transforms in potential energy is the kinetic energy 1 / 2 m V2 and the springs accept them before shortening too much. The kinetic energy depends on the mass of the cabin and the drop height. The potential energy of a compressed or stretched spring is 1 / 2 k L2 where k is the spring constant.(newtons/meter) and L is the variation in length of the spring.
Energy Equation,
1/2 m v2 = 1/2 k L2
0.5 * 1000 * 36 = 0.5 * k * 0.094 2
Spring Constant (k) = 36000 / 0.008836 = 4.1 x 10 7 N
The total energy of the elevator before it touches touches the spring is 0.5 * 1000 * 36 = 18000 Joules.
The elevator shows a stretch of Length by Sqrt ((2(- m * g * d + f) ) / k ) = x
Sqrt ( 2 (-1000 * 9.8 * 2.00+ 20000 ) / 4.1 x10 7 ) = 0.316 m
Initial Energy = Energy loss due to friction + Final Energy
Total Energy of the elevator, Initial Energy for displacement of 0.094m = m.g. ( H + L) = 1000 * 9.8 * (2.00 + 0.094) = 9.8 * 2094 = 20,521.2 Joules
Energy loss due to friction by the spring = m.g.x= Work done by the Frictional Force
=> 1000 * 9.8 * 0.316 = Work done by the Frictional Force .
=> 30,348 Joules
Total Energy of the Spring = 1/2 * k * x2
= 0.5 * 4.1 * 107 * 0.3162
= 2.05 * 0.316 * 0.316 * 10000000
= 2047 KJ
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