Physics 1 (Work and Energy) Please help, will rate. THANK YOU!!!! Starting from
ID: 1451188 • Letter: P
Question
Physics 1 (Work and Energy) Please help, will rate. THANK YOU!!!!
Starting from rest, a 2500 kg elevator accelerates upward at 1.20 m/s^2 for 8.00 m. How much work does gravity do on the elevator? How much work does the tension in the elevator cable do on the elevator? Write down Newton's Second Law for the elevator: Now calculate the work down by the tension in the elevator cable. Note the tension (force) of the cable is in the same direction as the displacement: Use the Work-Energy Theorem to find the kinetic energy of the elevator as it reaches 8.00 m. What is the speed of the elevator as it reaches 8.00 m?Explanation / Answer
Work = Force x distance (W = Fd)
A) Work done by gravity:
W = 2500(9.81)(8)
W = 196200
W = -1.96 x 10^5 J
B) Work done by tension:
Tension = mg + ma = m(g + a)
Tension = 2500(9.81 + 1.2) = 27,525 N
Work = Tension x distance
W = 27,525(8)
W = 220200
W = 2.20 x 10^5 J
C)
X = (1/2)*a*t^2
v = a*t or t = v/a
X = (1/2)*a*(v/a)^2
v^2 = 2*x*a
v = sqrt(2*X*a) = sqrt(2*8*1.2) = 4.38m/s
KE = (1/2)*m*v^2 = (1/2)*2500*4.38^2 = 23980.5 J
By Work-KE:
The work to overcome gravity is:
Wg = m*g*h = 2500*9.8*8 = 196000 J
The force to accelerate the elevator is:
F = m*a = 2500*1.2 = 3000
The work in accelerating the elevator is:
Wa = F*h = 2500*1.2*8 = 24000 J
Total work in the elevator cable is Wg + Wa = 220000 J
Only Wa contributes to the Kinetic Energy, Wg contributes to potential energy gain.
D) Change in kinetic energy = Total Work ((1/2)mv^2 = Fd)
(1/2)(2500)v^2 - 0 = -196200 + 220200
v = 4.38 m/s