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aniked [119]
3 years ago
8

A truck accelerating at 0.0083 meters/secondÆ covers a distance of 5.8 × 10Ê meters. If the truck's mass is 7,000 kilograms, wha

t is the work done to reach this distance?
Physics
2 answers:
laiz [17]3 years ago
8 0

Answer:

58.1*(5.8*10^4)  =3,369,800 Joules. hope it helps

tigry1 [53]3 years ago
3 0

work done = force * distance moved (in direction of the force)

force= mass* acceleration 

force=58.1N

58.1*(5.8*10^4)
=3,369,800 J
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Is there a frame of reference one can go into that seems to eliminate gravity as Newton described it?
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3 years ago
An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

8 0
3 years ago
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