# Work, Energy & Power Questions

Problem 1:

The dumbwaiter D and its load have a combined mass of 300kg, while the counterweight C has a mass of 400 kg. Determine the power delivered by the electric motor M when the dumbwaiter:

(a) is moving up at a constant speed of 2.5 m/s;

(b) has an instantaneous velocity of 2.5 m/s and an acceleration of 0.75 m/s^2, both directed upward.

Problem 2:

Two barges, each with a displacement (mass) of 500 Mg, are loosely moored in calm water. A stunt driver starts his 1500 kg car from rest at A, drives along the deck, and leaves the end of the 15°ramp at a speed of 50 km/h relative to the barge and ramp. The driver successfully jumps the gap and brings his car to rest relative to barge 2 at B.

Calculate the velocity v2 imparted to barge 2 just after the car has come to rest on the barge. Neglect the resistance of the water to motion at the low velocities involved.

Problem 3:

The figure shows the drop tower in Dream World at Gold Coast. The major dimensions are shown in the figure. The carriage mass is 1523 kg and carries a maximum number of 8 riders with an average of 75 kg per person. The vehicle drops from the rest position at a height of 110000mm until the position at 35000mm from the ground level. After this position, constant brakes are applied to slow the vehicle down. Assuming there is no friction force along the track:

1) Calculate the velocity at the position where the brakes start to engage;

2) Calculate the deceleration during braking; and

3) Using the energy method, calculate the brake force and the energy required to bring the carriage to a stop (900mm above ground level).

Please see attached file for figures.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please refer to the attachment.

Solution 1: Pulley 2 Pulley 3

Cord 1

T1 Cord 2

T1 T1 T2 T2

Pulley 1 T1 T2

C D

T2

400x9.8=3920 N 300x9.8=2940 N M

Let tensions in two cords be T1 and T2 as shown.

If the motor pulls in a length x of the cord 2, mass D moves up by the same distance x and consequently cord 1 attached to D moves up by distance x. However, mass C moves down by a distance x/2 as length x of the cord gets equally divided on either side of pulley 1.

Hence, speed of C is one half of the speed of D in the opposite direction and acceleration of C is one half of that of D.

Net upward force on D = FD = (T1 + T2) - 2940

Net downward force on C = FC = 3920 - 2T1

a) Free body diagram of D is drawn below:

T1 T2

D

2940 N

As D is moving up at constant speed i.e. without acceleration, net force on it must be zero. Hence, (T1 + T2) - 2940 = 0

(T1 + T2) = 2940 ........(1)

Free body diagram of C as as follows:

T1 T1

C

3920 N

Similarly, net force on C must be zero. Hence, 3920 - 2T1 = 0, T1 = 1960 N (↑)

Substituting for T1 in (1): T2 = 2940 - 1960 = 980 N (↑)

We know that Power P = Force x Velocity. As the motor is pulling in the cord 2 against a tension of 980 N at a speed of 2.5 m/s, motor is delivering power P = 980 x 2.5 = 2450 watt

b)

T1 T2

D a

2940 N

D is moving up at constant acceleration a = 0.75 m/s2. Acceleration a = Net force/Mass. Hence, [(T1 + T2) - 2940]/300 = 0.75

(T1 + T2) = 3165 ........(2)

Similarly, for C:

T1 T1

C a/2

3920 N

(3920 - 2T1)/400 = 0.75/2 (acceleration of C is ½ of acceleration of D)

T1 = 1885 N (↑)

Substituting for T1 in (2): T2 = 3165 - 1885 = 1280 N (↑)

Power delivered by the motor P = 1280 x 2.5 = 3200 watt

Solution 2: This problem can be solved on the basis of conservation of linear momentum.

Mass of barge 1 = Mass of barge 2 = M = 500 x 106 g or 500 x 103 kg

Mass of the car = m = 1500 kg

Velocity of car as it leaves barge 1 = v = 50 km/hr = 50000/3600 = 13.89 m/s at an angle 15O with the horizontal. (OR 13.89 m/s / 15O)

Horizontal component of the velocity = vcos15O = 13.89 x 0.966 = 13.42 m/s (→)

We will designate velocity of car relative to the barge as VCB and that relative to water as VCW. Also we will designate velocity of barge 1 relative to water as VB1W (similarly for barge 2 VB2W)

As the system of car and barge is an isolated system, conservation of linear momentum holds. Using conservation of momentum at (→)x direction, M x VB1W + m x VCW = 0 (as initially the system is at rest)

Substituting values we get: 500000VB1W + 1500VCW = 0

333.33VB1W + VCW = 0 ........(1)

As car's horizontal velocity component relative to the barge is known (13.42 m/s), we can write the following equation: VCB1 = VCW - VB1W = 13.42 (→) .........(2)

Subtracting (2) from (1): 333.33VB1W + VCW - VCW + VB1W = - 13.42

334.33VB1W = - 13.42

VB1W = - 0.04 m/s (→) = 0.04m/s (←)

Substituting in (2): VCW = 13.42 + VB1W = 13.42 - 0.04 = 13.38 m/s(→)

As the car lands on barge 2, it has a momentum of 1500 x 13.38 = 20000 kg.m/s

Initial momentum of barge 2 is zero. Hence, net momentum of barge 2 and car just before the car lands on barge 2 = 20000 kg.m/s →

Final velocity of car on barge 2 = 0.

Momentum of the barge 2 and car is given by: 500000VB2W + 1500VCW

By conservation of momentum: 500000VB2W + 1500VCW = 20000

333.33VB2W + VCW = 13.33 .........(3)

As finally velocity of car relative to barge 2 is zero, we can write the following equation:

VCB2 = VCW - VB2W = 0 .........(4)

Subtracting (4) from (3): 333.33VB2W + VCW - VCW + VB2W = 13.33

334.33VB2W = 13.33

VB2W = 0.0399 m/s or 3.99 cm/s (→)

Immediately after car comes to rest, barge 2 moves forward with a speed of 3.99 cm/s.

Solution 3: 1) As the carriage (with riders) starts from rest, its initial velocity u = 0

During the initial fall, the carriage falls freely. Hence, its acceleration is acceleration due to gravity (a = g = 9.8 m/s2 ↓).

Displacement of the carriage during the free fall period = s = 110 - 35 = 75 m ↓

We can apply the kinematic equation v2 - u2 = 2as where v = final velocity

v2 - 0 = 2 x 9.8 x 75 = 1470

v = 38.34 m/s ↓

2) During the next period of braking, the carriage undergoes deceleration.

The initial velocity for this phase u' = 38.34 m/s. ↓

Final velocity for this phase v' = 0 (as the carriage finally comes to rest).

Displacement s' = 35 - 0.9 = 34.1 m↓

Again applying the equation v'2 - u'2 = 2as'

0 - 38.342 = 2 x a x 34.1

Acceleration a = - 21.55 m/s2 = 21.55 m/s2 ↑ (negative acceleration implies deceleration)

3) Kinetic energy at the end of free fall period = KE = ½ mv2 where m = mass of the carriage and riders = 1523 + 8 x 75 = 2123 kg

KE = ½ x 2123 x 38.342 = 1560400 J

Potential energy of the carriage (with respect to its lowest position) = PE = mgh = 2123 x 9.8 x 34.1 = 709464 J

Total mechanical energy possessed by the carriage = 1560400 + 709464 = 2269864 J

This energy is dissipated by the friction applied by the brakes. In other words, the work done by the frictional force of brakes equals this energy.

Hence, work required to be done (energy required) to bring the carriage to rest = 2269864 J

Let the average braking force be F. Displacement during braking = 34.1 m

(that is, Displacement of the brake shoe = Displacement of the carriage during braking period)

Work done = Force x Displacement = F x 34.1 = 2269864 J

F = 66565 N

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