NAS 236 Quiz 4 Study Guide

Pataasin ang iyong marka sa homework at exams ngayon gamit ang Quizwiz!

The second floor of a house is 6 m above the street level. How much work is required to lift a 300-kg refrigerator to the second-story level?

Work 5 DE 5 Dmgh 5 300 kg 3 10 N/k 3 6 m 5 18,000 J. 43. At three times the speed, it has nine times (32 ) the KE and will skid nine times as far—135 m. Since the frictional force is about the same in both cases, the distance has to be nine times as great for nine times as much work done by the pavement on the car.

Show that 50 W of power is required to give a brick 100 J of PE in a time of 2 s.

100 J/2 s = 50 Watts

If you push a crate horizontally with 100 N across a 10-m factory floor and the friction between the crate and the floor is a steady 70 N, how much kinetic energy does the crate gain?

300 J

In the hydraulic machine shown, you observe that when the small piston is pushed down 10 cm, the large piston is raised 1 cm. If the small piston is pushed down with a force of 100 N, what is the most weight that the large piston can support?

48.(F x d)in = (F x d)out(100 N x 10 cm)in = (? x 1 cm)outSo we see that the output force and weight held is 1000 N (less if efficiency < 100%).

A moving hammer hits a nail and drives it into a wall. If the hammer hits the nail with twice the speed, how much deeper will the nail be driven? If the hammer hits with three times the speed?

According to the work-energy theorem, twice the speed corresponds to 4 times the energy, and therefore 4 times the driving distance. At 3 times the speed, driving distance is 9 times as much.

Consider the swinging-balls apparatus. If two balls are lifted and released, momentum is conserved as two balls pop out the other side with the same speed as the released balls at impact. But momentum would also be conserved if one ball popped out at twice the speed. Discuss why this never happens. (And explain why this exercise is in Chapter 7 rather than in Chapter 6.)

But now look at this situation in accordance with the conservation of energy (½ m v2)If only 1 ball pops out with twice the velocityEnergy before = ½ m v2 + ½ m v2 = m v2Energy after = ½ (m) (2 v)2= ½ (m) (4 v2) = 2 m v2These energies are not equal and, therefore, the law of conservation of energy is not obeyed!But, since 2 balls actually popped outEnergy before = ½ m v2 + ½ m v2 = m v2Energy after = ½ m v2 + ½ m v2 = m v2These energies are equal and, therefore, both laws of conservation are obeyed.

An apple hanging from a limb has potential energy because of its height. If it falls, what becomes of this energy just before it hits the ground? When it hits the ground?

Immediately before the apple hits the ground, its initial PE becomes KE. When it hits the ground its energy becomes thermal energy.

When the speed of a moving car is doubled, how much more kinetic energy does it have?

It has four times as much.

Which requires more work: lifting a 50-kg sack a vertical distance of 2 m or lifting a 25-kg sack a vertical distance of 4 m?

It is the same, for the product of each is the same; (50 kg) (2 m) = (25 kg) (4 m).

A car is raised a certain distance in a service-station lift and therefore has potential energy relative to the floor. If it were raised twice as high, how much more potential energy would it have?

It would have twice the PE because the distance is two times greater.

Show that the gravitational potential energy of a 1000-kg boulder raised 5 m above ground level is 50,000 J. (You can express g in units of N/kg because m/s2 is equivalent to N/kg.)

PE 5 mgh 5 (1000 kg) (10 N/kg)(5 m) 5 50,000 Nm 5 50,000 J.

How many watts of power do you expend when you exert a force of 50 N that moves a crate 8 m in a time interval of 4 s?

Power 5 Fd/t 5 (50 N)(8 m)/(4 s) 5 100 J/1 s 5 100 watts.

On a playground slide, a child has potential energy that decreases by 1000 J while her kinetic energy increases by 900 J. What other form of energy is involved, and how much?

The 100 J of potential energy that doesn't go into increasing her kinetic energy goes into thermal energy—heating her bottom and the slide.

At what point in its motion is the KE of a pendulum bob at a maximum? At what point is its PE at a maximum? When its KE is at half its maximum value, how much PE does it have relative to its PE at the center of the swing?

The KE of a pendulum bob is maximum where it moves fastest, at the lowest point; PE is maximum at the uppermost points. When the pendulum bob swings by the point that marks half its maximum height, it has half its maximum KE, and its PE is halfway between its minimum and maximum values. If we define PE = 0 at the bottom of the swing, the place where KE is half its maximum value is also the place where PE is half its maximum value, and KE = PE at this point. (By energy conservation: Total energy = KE + PE.)

When the mass of a moving object is doubled with no change in speed, by what factor is its momentum changed? By what factor is its kinetic energy changed?

When the mass is doubled with no change in speed, both momentum and KE are doubled.

Two people who weigh the same climb a flight of stairs. The first person climbs the stairs in 30 s, and the second person climbs them in 40 s. Which person does more work? Which uses more power?

Work done by each is the same, for they reach the same height. The one who climbs in 30 s uses more power because work is done in a shorter time.

Show that the kinetic energy of a 1.0-kg book tossed across the room at a speed of 3.0 m/s is 4.5 J. (1 J is equivalent to 1 Joule 5 1.0-kg *m2 /s2 .)

m = 1, kgv = 3 m/s1/2, (1 kg)(3 m/s)2 = 4.5 J, 1/2 (9) = 4.5 J

Calculate the work done when a force of 5 N moves a book 1.2 m. (Remember that 1 N?m 5 1 J.)

(5N)(1.2m) = 6.0 J

(a) How much work is done when you push a crate horizontally with 100 N across a 10-m factory floor? (b) If the force of friction on the crate is a steady 70 N, show that the KE gained by the crate is 300 J. (c) Show that 700 J is turned into heat.

(a) You do F d = 100 N 10 m = 1000 J of work.(b) Because of friction, net work on the crate is less. KE = net work = net force distance = (100 N - 70 N)(10 m) = 300 J(c) So the rest, 700 J, goes into heating the crate and floor.


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