Ask question

Ask question

All categories

3 years ago

10

A 4.50-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.650 kg ,

initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190 m/s. The block slides a distance of 72.0 cm along the surface from its initial position.
a. What is the coefficient of kinetic friction between block and surface?
b. What is the decrease in kinetic energy of the bullet?
c. What is the kinetic energy of the block at the instant after the bullet passes through it?

1 answer:

Citrus2011 [14]3 years ago

5 0

Answer:

a) μ = 0.1957 , b) ΔK = 158.8 J , c) K = 0.683 J

Explanation:

We must solve this problem in parts, one for the collision and the other with the conservation of energy

Let's find the speed of the wood block after the crash

Initial moment. Before the crash

p₀ = m v₁₀ + M v₂₀

Final moment. Right after the crash

pf = m $v_{1f}$ + M v_{2f}

The system is made up of the block and the bullet, so the moment is preserved

p₀ = pf

m v₁₀ = m v_{1f} + M v_{2f}

v_{2f} = m (v₁₀ - v_{1f}) / M

v_{2f} = 4.5 10-3 (400 - 190) /0.65

v_{2f} = 1.45 m / s

Now we can use the energy work theorem for the wood block

Starting point

Em₀ = K = ½ m v2f2

Final point

Emf = 0

W = ΔEm

- fr x = 0 - ½ m v₂₂2f2

The friction force is

fr = μN

With Newton's second law

N- W = 0

N = Mg

We substitute

-μ Mg x = - ½ M v2f2

μ = ½ v2f2 / gx

Let's calculate

μ = ½ 1.41 2 / 9.8 0.72

μ = 0.1957

b) let's look for the initial and final kinetic energy

K₀ = 1/2 m v₁²

K₀ = ½ 4.50 10⁻³ 400²

K₀ = 2.40 10² J

Kf = ½ 4.50 10⁻³ 190²

Kf = 8.12 10¹ J

Energy reduction is

K₀ - Kf = 2.40 10²- 8.12 10¹

ΔK = 158.8 J

c) kinetic energy

K = ½ M v²

K = ½ 0.650 1.45²

K = 0.683 J

You might be interested in

A force of 8,480 N is applied to a cart to accelerate it at a rate of 26.5 m/s2. What is the mass of the cart?

Anni [7]

F=ma
8480=26.5m
m=8480/26.5
m=320
The mass of the cart is 320kg.

8 0

3 years ago

Read 2 more answers

A healthy astronaut's heart rate is 60 beats/min. Flight doctors on Earth can monitor an astronaut's vital signs remotely while

Anton [14]

GIVEN:
60 beats per minute
21 beats per minute
find x= how fast would an astronaut be flying away
1 x
----- * ------ = (60x = 21) -------> 60x = 21 ------------> x= 0.35
60 21 ------- -----
60 60

The answer is 0.35 seconds which refers to how fast would an astronaut be flying away from the earth if he has a heart rate of 21 beats/min.

5 0

3 years ago

Read 2 more answers

Does anyone know this

NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the right

Explanation:

1)

To find the net force, we have to analyze all the forces acting on the box.

We have:

Force to the right: $F_a = 20 N$ , the applied force
Force to the left: $F_f = 4 N$ , the force of friction
Force to the bottom: $F_g = 400 N$ , the weight of the box (the weight is always downward vertically)
Force to the top: $F_N = 400 N$ . This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

$F_x = F_a - F_f = 20 - 4 = 16 N$ (to the right)

While the vertical force is

$F_y = F_N - F_g = 400 - 400 = 0$

So the net force is 16 N to the right.

2)

In this case, we have the following forces:

$F_g = 4 N$ downward, the weight of the ball
$F_a = 100 N$ to the left, the force that kicks the ball
$F_f = 2 N$ to the right, the force of friction
$F_N = 4 N$ upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

$F_x =F_a - F_f = 100 -2 = 98 N$ (to the left)

While the vertical force is

$F_y = F_g - F_N = 4 - 4 = 0$ (downward)

And so, the net force is 98 N to the left.

3)

The force acting on the squirrel in this problem are:

$F_g = 8 N$ downward, the weight of the squirrel
$F_f = 7.5 N$ upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

$F_y = F_g - F_f = 8 - 7.5 = 0.5 N$

And since the weight is larger than the air resistance, the direction of the net force is downward.

4)

The forces acting on Monkey are:

$F_1=95 N$ is the force applied to the right by Bunny
$F_2 = 75 N$ is the force applied by Deer from the left (so, also on the right)
$F_g = 50 N$ is the weight of Monkey, downward
$F_N = 50 N$ is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

$F_x = F_1 + F_2 = 95+75=170 N$ (to the right)

While the net force in the vertical direction is

$F_y = F_N - F_g = 50 - 50 = 0$

And therefore the net force is 170 N to the right

5)

The forces acting on Deer are:

$F_a = 100 N + 100 N = 200 N$ to the right, the combined force applied by Bunny and Monkey
$F_f = 25 N$ to the left, the force of friction
$F_g = 150 N$ downward, the weight of the deer
$F_N = 150 N$ upward, the normal reaction from the surface that balances the weight

So the net horizontal force is

$F_x = F_a - F_f = 200 - 25 = 175 N$ to the right

While the net vertical force is

$F_y = F_N - F_g = 150 - 150 = 0$

So the net force is 175 N to the right.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

3 0

3 years ago

Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o

Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

$\Sigma F = F - F' = M\cdot a$

Box with mass 2M

$\Sigma F = F' - F'' = 2\cdot M \cdot a$

Box with mass 3M

$\Sigma F = F'' = 3\cdot M \cdot a$

On the third equation, acceleration can be modelled in terms of F'':

$a = \frac{F''}{3\cdot M}$

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

$F' = 2\cdot M \cdot a + F''$

$F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''$

$F' = \frac{5}{3}\cdot F''$

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

$F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)$

$F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''$

$F = 2\cdot F''$

$F'' = \frac{1}{2}\cdot F$

Afterwards, F' as function of the external force can be obtained by direct substitution:

$F' = \frac{5}{6}\cdot F$

The net forces of each block are now calculated:

Box with mass M

$M\cdot a = F - \frac{5}{6}\cdot F$

$M\cdot a = \frac{1}{6}\cdot F$

Box with mass 2M

$2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F$

$2\cdot M \cdot a = \frac{1}{3}\cdot F$

Box with mass 3M

$3\cdot M \cdot a = \frac{1}{2}\cdot F$

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0

3 years ago

NikAS [45]

Acceleration is maximum. <span>Velocity is decreasing. Acceleration is increasing</span>

7 0

3 years ago

Read 2 more answers