How much kinetic energy does a 3200.0 kg missile moving at 250.0 m/s have? a. 4.

Question

How much kinetic energy does a 3200.0 kg missile moving at 250.0 m/s have? a. 4.6 times 10^5 J b. 1.0 times 10^5 J c. 2.0 times 10^5 J d. 9.2 times 10^5 J A 17.2 kg ball is held 15.0 m above the ground. What is its gravitational potential energy? a. 258.0 J b. 168.6 J c. 2528.4 J d. 3947.3 J What is the mechanical energy of 0.3 kg arrow shot at 100.0 m/s from 1.5 m above the ground? a. 4.4 J b. 0.0 J c. 1500.0 J d. 1504.4 J In the absence of nonconservative forces, is the mechanical energy of an object conserved? a. Yes b. No A spring with spring constant k = 1000.0 N/m is compressed a distance of x = 0.5 m from equilibrium. How much energy is stored in it? a. 500.0 J b. 125.0 J c. 250.0 J d. 1000.0 J Ignoring air resistance, how fast would a box hit the ground if dropped from 28.9 m? a. 16.8 m/s b. 23.8 m/s c. 566.4 m/s d. 283.2 m/s A 1500.0 kg Tonka Truck going 30.5 m/s hits a spring which compresses 5.4 m before the Tonka Truck stops. What is the spring constant? a. 8472.2 N/m b. 129201.4 N c. 75238.3 N/m d. 47852.4 N/m

Explanation / Answer

1.)

Kinteic energy is,

KE = (1/2)(3200 kg)(250 m/s)2

or, KE = 1.0X108 J

So, the correct option is b.

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2.)

Graviational potential energy is,

PE = mgh = (17.2 kg)(9.8 m/s2)(15 m)

or, PE = 2528.4 J

So, the correct option is c.

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3.)

Mechanical energy ME = potential energy + kinetic energy

or, ME = (0.3 kg)(9.8 m/s2)(1.5 m) + (1/2)(0.3 kg)(100 m/s)2

or, ME = 1504.41 J

So, the correct option is d.

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4.)

In absence of non-conservative forces, the mechanical energy is conserved.

So correct option is a.

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5.)

Energy stored = (1/2)kx2

or, Energy stored = (0.5)(1000 N/m)(0.5 m)2

or, Energy stored = 125 J

So, the correct option is b.

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6.)

Initial energy of the block is PE = mgh

final energy is KE = (1/2)mv2

So from conservation of energy we have,

(1/2)mv2 = mgh

or, v2 = 2gh = 2(9.8 m/s2)(28.9 m)

or, v = 23.8 m/s

So, the correct option is b.

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7.)

Initial energy of the spring+truck system is KE = (1/2)mv2

Final energy of the spring+truck system is PE = (1/2)kx2

So, from conservation of energy we have,

(1/2)mv2 = (1/2)kx2

or, k = mv2/x2 = (1500 kg)(30.5 m/s)2/(5.4 m)2

or, k = 47852.36 N/m

So, the correct option is d.

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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