Stacy's home is sprayed with a pesticide to control spiders. She didn't see any spiders for a period of time, but eventually, sh
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Answer:
(a). The acceleration is 3.20 m/s².
(b). The amount of work done by each force is 19.2 J.
(c). The total work done on the block is 19.2 J.
(d). The final speed of the block is 6.26 m/s.
Explanation:
Given that,
Mass of block = 3 kg
Distance = 2 m
Angle = 30°
Coefficient of kinetic friction = 0.20
(a). We need to calculate the acceleration
Using balance equation of force
Put the value into the formula
The acceleration is 3.20 m/s².
(b). We need to calculate the amount of work done by each force
Using formula of work done
Normal force is
So due to this the net force is zero then the no work done by reaction force.
By another force,
Put the value into the formula
The amount of work done by each force is 19.2 J.
(c). We need to calculate the total work done on the block
The total work done on the block is 19.2 J.
(d). We need to calculate the final speed of the block
Using equation of motion
Put the value into the formula
The final speed of the block is 6.26 m/s.
Hence, This is the required solution.
Answer:
a) 3.43 m/s
Explanation:
Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.
The total momentum before the bullet is shot is zero, because they are both at rest, so:
Instead the total momentum of the system after the shot is:
where:
m = 0.006 kg is the mass of the bullet
M = 1.4 kg is the mass of the rifle
v = 800 m/s is the velocity of the bullet
V is the recoil velocity of the rifle
The total momentum is conserved, therefore we can write:
Which means:
Solving for V, we can find the recoil velocity of the rifle:
where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is
a) 3.43 m/s
<u>Answers:</u>
In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:
<h2>This can be read as: The Net Force of an object is equal to its mass
multiplied by its acceleration
.
We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension of the string (<u>See figure attached). </u>
We already know<u> is greater than
</u>, this means the weight of the block 2
is greater than the weight of the block 1
; therefore <u>the acceleration of the system will be in the direction of
</u>, as shown in the figure attached.
We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.
This means that the tension will be the same along the string regardless of the difference of mass of the blocks.
Now that we have the conditions clear, let’s begin with the calculations:
1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.
This is done using equation (1), where the force of the weight is calculated using the <u>acceleration of gravity</u>
acting on the blocks:
<h2>
<u>For block 1: </u>
(3)
<u>For block 2: </u>
(5)
Then, we are going to <u>find the acceleration of the whole system:
</u>
(7)
Where the Resulting Force is equal to the sum of the weights
and
.
In the figure attached, note that is in opposite direction to the acceleration
, this means it must <u>have a negative sing</u>; while
is in the same direction of
.
Here we only have to isolate from equation (8) and substitute the values according to the conditions of the system:
Then:
2) For the second part of the problem, we have to find the tension of the string.
We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.
Let’s prove it:
For
we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:
(9)
(10)
Then;
<h2>For
we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:
(9)
(10)
Then;
<h2>