Classify the following situations as involving balanced or unbalanced forces.

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

3 0

4 years ago

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Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w

Scilla [17]

Answer:

d= 5.62 m

Explanation:

In order to have destructive interference, the path difference from the sources to the listener must be an odd multiple of half wavelengths, as follows:

d = (2n+1) * λ/2

In orfer to know which is the wavelength, we can use the relationship between propagation speed (in this case speed of sound), frequency and wavelength:

v= λ*f ⇒ λ = v/f = 344 m/s / 688 1/sec = 0.5 m

So, the path difference must be, at least, λ/2:

d = b-a = λ/2, where b is the distance to the speaker B, and a, the distance to the speaker A.

Applying Pithagorean Theorem, as the perpendicular distance d (which is our unknown) is the same for the triangles defined by the horizontal distance to the listener, and the straight line from the new position to the sources, we can write:

d² = a²- (3.0)²

d² = b²- (3.5)²

As the left sides are equal, so do right sides:

a² - (3.0)² = b² - (3.5)²

⇒ b² - a² = (3.5)² - (3.0)² = 3.25

We can replace (b²- a²) as follows:

b² - a² = (b+a)(b-a) = 3.25 (2)

We know that b-a = λ/2 = 0.25

Replacing in (2), we have:

b+a = 3.25 / 0.25 = 13 m

b-a = 0.25 m

Adding both sides:

2*b = 13.25 m ⇒ b= 13.25 /2 = 6.63 m

⇒ d² = (6.63)² - (3.5)² = 31.6 m²

⇒ d=√31.6 m² = 5.62 m

3 0

3 years ago

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same am

Tcecarenko [31]

Answer:

Incomplete question,

This is the complete question

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration ~a of the block after it begins to move. Express your answer in terms of some or all of the variables µs, µk, and m, as well as the acceleration due to gravity g.

Explanation:

Let the force that make the object to start moving be F,

Frictional force is opposing the motion, the body has to overcome two frictional forces acting in the opposite direction of the motion.

Also, weight and normal reaction are acting in vertical direction, the weight is acting downward while the reaction is acting upward.

Weight of the object is given as

W=mg

Analyzing the vertical motion i.e y-axis.

ΣF = ma

since the body is not moving upward, the a=0

N-W=0

Then, N=W

So, N=mg