Mechanics

a) i) Since there is no acceleration, the resultant force should be zero.
tension to the car should be 750 + 6500 = 7250 N
ii) The same reason , tension = 600 N
iii) The difference in the 2 tension is balanced by friction of the drum.

圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/A_Hepburn01Feb201823.jpg?t=1203503060


Denote in the figure:

T1 be the tension in the part of the cable connected to the car

T2 be the tension in the part of the cable connected to the counterweight

mcg be the weight of the counterweight

mlg be the weight of the car

mpg be the weight of the passenger

a. As the system is moving in uniform velocity

By Newton’s 1st law of motion, the net force acting on the objects should be zero.

i. T1 – (mlg + mpg) = 0

T1 = mlg + mpg

T1 = 6500 + 750

T1 = 7250 N

ii. T2 = mcg

T2 = 600 N

iii. The two tensions are different from each other since frictional force exists between the drum and the cable.

b.i. Gain in height of the passenger in one second, h = ut = 2 m

Useful power to raise the passenger

= Gain in P.E. / time

= mpgh / t

= (750)(2) / (1)

= 1500 W

ii. Power lost

= Work done against friction / time

= (T1 – T2)h / t

= (7250 - 600)(2) / 1

= 13 300 W

c. The weight of the couterweight is to balance the weight of the car.

d. A frictionless drum cannot be used since frictional force is necessary for the raising of the car.