A 180kg load is moved 23m vertically with an accelaration..

Question

A 180kg load is moved 23m vertically with an acceleration a=0.150g by a single cable.
Determine:
a. the tension in the cable
b. the net work done on the load
c. the work done by the cable on the load
d. the work done by the gravity on the load
e. the final speed of the load assuming it started from rest

 

Answers: a) 2029 N, b) 6086 J, c) 46658 J, d) -40600 J, e) 8.22m/s

Answer 1

(a)

 

T - mg = ma

T = m(a + g) = (180 kg) (9.8 m/s^2 + 0.150 * 9.8 m/s^2) = 2028.6 N

 

(b)

 

W = Fd = (T - mg)d = (2028.6 N - 180 kg * 9.8 m/s^2) (23 m) = 6085.8 J

 

(c)

 

W_c = Td = (2028.6 N) (23 m) = 46,657.8 J

 

(d)

 

W_g = -mgd = -180 kg * 9.8 m/s^2 * 23 m = -40,572 J

 

(e) 

 

W = delta KE = KE_f - KE_i = KE_f - 0 = 1/2 mv^2

v = sqrt(2 W / m) = sqrt (2 * 6085.8 / 180) m/s = 8.22 m/s