The Isolated System—Conservation of Mechanical Energy Problems and Solutions 2

 Problem#1

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 30.0° above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 s and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? (b) What maximum power must the winch motor provide? (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1 250 m?

 

Answer:

The energy of the car is

E = ½ mv2 + mgy = ½ mv2 + mgd sinθ

where d is the distance it has moved along the track.

Then, the power of the car given by

P = dE/dt = mv(dv/dt) + mgvsinθ

(a) Power must the winch motor provide when the car is moving at constant speed (dv/dt = 0) is

P = mgvsinθ = 950 kg(9.80 m/s2)(2.20 m/s)sin30.00 = 10.2 kW

(b) Maximum power is injected just before maximum speed is attained:

With a = ∆v/∆t = (2.20 m/s2 – 0)/(12 s) = 0.183 m/s2

Then,

P = (950 kg)(2.20 m/s)(0.183 m/s2) + (1.02 x 104 W) = 10.6 kW

(c) At the top end,

½ mv2 + mgdsinθ = (950 kg)[½(2.20 m/s)2 + (9.80 m/s2)(1250 m) sin30.00] = 5.82 MJ

 

Problem#2

A simple pendulum, which we will consider in detail in Chapter 15, consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is 2.00 m long and makes an initial angle of 30.0° with the vertical, calculate the speed of the particle (a) at the lowest point in its trajectory and (b) when the angle is 15.0°.

Answer:

(a) Energy of the object-Earth system is conserved as the object moves between the release

point and the lowest point. We choose to measure heights from y = 0 at the top end of the

string.

we use

Ui + Ki = Uf + Kf

mgyi + 0 = mgyf + ½ mvf2

gdcosθ = gd + ½ vf2

vf2 = 2gd(1 – cosθ)

vf2 = 2(9.80 m/s2)(2.00 m)(1 – cos30.00)

vf = 2.29 m/s

 

(b) the speed of the particle when the angle is 15.0°.

Choose the initial point at 30.00 and the final point at 15.00:

mg(–Lcos30.00) + 0 = mg(–Lcos15.00) + ½ mvf2

vf2 = 2gL(cos15.00 – cos30.00)

vf2 = 2(9.80 m/s2)(2.00 m)(cos15.00 – cos30.00)

vf = 1.98 m/s

 

Problem#4

An object of mass m starts from rest and slides a distance d down a frictionless incline of angle &. While sliding, it contacts an unstressed spring of negligible mass as shown in Figure P8.10. The object slides an additional distance x as it is brought momentarily to rest by compression of the spring (of force constant k). Find the initial separation d between object and spring.

Answer:

Choose the zero point of gravitational potential energy of the object-spring-Earth system as the

configuration in which the object comes to rest. Then because the incline is frictionless, we have

EA = EB

KB + UgB + UsB = KA + UgA + UsA

0 + mg(d + x) cosθ + 0 = 0 + 0 + ½ kx2

d + x = (½ kx2)/(mgcosθ)

then

d = (kx2/2mgcosθ) – x

Problem#5

A block of mass 0.250 kg is placed on top of a light vertical spring of force constant 5 000 N/m and pushed downward so that the spring is compressed by 0.100 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Answer:

From conservation of energy for the block-spring-Earth system,

Ki + Ugi + Usi = Kf + Ugf + Usf

0  + mgh + 0 = 0 + 0 + ½ kx2

(0.250 kg)(9.80 m/s2)h = ½ (5000 N/m)(0.100 m)2

This gives a maximum height h = 10.2 m 

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