Energy Review

 

1. A passenger airplane weighing 2.4 X 10^6 N is traveling at 256 m/s. Determine it's kinetic energy.

Scroll down for answer First conver weight to mass: (2.4 X 10^6 kg) / (9.8 m/s/s) = 2.45 X 10^5 kg Then 1/2 mv^2 = 1/2 (2.45 X 10^5)(256)^2 = 8.03 X 10 ^9 J

 

2. A 956 kg car accelerates from 20.0 m/s to 55 m/s in order to pass a car on the freeway.

a.) What were the cars initial and final KE?

b.) How much work was required for the acceleration to occur?

Scroll down for answer K = 1/2 mv^2 therefore K1 = (1/2 (956-kg)(20.0 m/s)^2 = 1.91 X 10^5 J K2 = 1/2 (956-kg)(55.0 m/s)^2 =  1.45 X 10^6 J So : W = K2 - K1 = 1.45 X 10^6 J - 1.91 X 10^5 J = 1.25 X 10^6 J

 

3. A firefighter with a mass of 55 kg climbs up a ladder 8 m from the base of a building. How much potential energy does the firefighter gain?

Scroll down for answer PE = mgh = (55 kg)(9.8 m/s/s)(8m) = 4,312 J

 

4. A student lifts a 10 kg box from a shelf 3 m high to a shelf 7 m high. What is the increase in the box's potential energy?

Scroll down for answer PE = mgh2 - mgh1 = mg delta h therefore (10 kg)(9.8 m/s/s)(7 m - 3 m) = 392 J

 

5. An 85 kg student takes a ride on a skateboard with a mass of 3 kg. The skateboard and the student 

have a combined KE of 550 J. How fast is the student traveling?

Scroll down for answer KE = 1/2 mv^2 therefore v = sq rt 2 KE/m so:  v = sq rt (2)(550 J) / 88 kg = 3.54 m/s  

 

6. A 60 kg skateboarder starts from rest at the top of a 50 m high slope. Allowing only the force of gravity to propel her ...

a.) What is her gravitational PE with respect to level ground at the bottom of the hill?

b.) What speed will she attain at the bottom of the hill? (and ignoring any resistances ...)

c.)  What energy is transferred to friction is she reaches the end of the slope at 20 m/s

Scroll down for answer a.) PE = mgh = (60 kg)(9.8 m/s/s)(50 m) = 2.9 X 10^4 J b.) With friction neglected, KEf = PEi  or delta KE = PEi vi = 0 and u = 0 therefore to solve for vf;  KE = 1/2 mv^2 becomes vf = sq rt 2 (KE) / m = sq rt 2 (2.94 X 10^4) / 60 kg = 31.30 m/s c.) Wf = mghi - 1/2 mvf^2 = (60 kg)(9.8 m/s/s)(50 m) - 1/2 (60 kg)(20 m/s)^2 therefore 29,400 J - 12,000 J = 1.7 4 X 10^4 J transferred to friction

 

7. A 25 kg cannon ball is fired straight up and reaches a maximum height of 85 m. a.) Determine the initial velocity of the ball.

Scroll down for answer In this situation the PE max = KE initial (law of conservation of energy) since KE and PE are scalar quantities no sign need be applied therefore: 1/2 mv^2 = mgh to solve for v mass cancels out and you get 1/2 v^2 = gh and  v^2 = 2gh then v = sq rt 2gh therefore v = sq rt 2gh = sq rt 2(9.8 m/s/s)(85 m) = 40.82 m/s or 4.08 X 10^1 m/s

8. Suppose a baseball 0.50 kg were thrown straight up and reached the same height ... what can you say about the baseballs initial velocity and why.

Scroll down for answer Since mass cancels out the initial velocity for any similar object would be the same.

 

And now for one of your favorites to sink your intellectual teeth into.

10. A 12 kg crate sits on the floor. We want to load it into a truck by sliding it up a ramp 3.0 m long, inclined 

at 30 degrees. An apprentice to the job, does not think about friction, calculates that he can get the crate up 

the ramp by giving it an initial speed of 5.0 m/s at the bottom and letting it go. But friction is not negligible; 

the crate slides 1.6 m up the ramp, stops, and slides back down. At the 1.6 m mark the height of the ramp 

is 0.80 m above the ground. Assuming that the frictional force is constant. 

Find the magnitude of the frictional force. 

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