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Valentin [98]
3 years ago
12

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

e horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

Physics
1 answer:
zhannawk [14.2K]3 years ago
6 0

Answer: V_{f}=2.96m/s    

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight w of the block has an x-component and y-component:

w_{x}=wsin(\theta)    (1)

w_{y}=wcos(\theta)    (2)

As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

N_{y}=Ncos(\theta)    (4)

In addition, we know N=w, then \sum F_{y}=0

In the X-component:

\sum F_{x}=m.a

m.a=w_{x}    (5)

Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

So:

m.g.sin(\theta)=m.a   (7)

a=g.sin(\theta)    (8)

a=5.88m/{s}^{2}    (9)   >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

d=\frac{1}{2}a{t}^{2}   (10)

We already know the value of  d and calculated a, we have to find t:

t=\sqrt{\frac{2d}{a}}   (11)

t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

If V_{o}=0

V_{f}=a.t   (14)

V_{f}=(5.88m/{s}^{2})(0.50s)   (15)

Finally we get the value of the Final Velocity of the block:

V_{f}=2.96m/s    

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The initial velocity of car 1 is 20 m/s to the right

The initial velocity of car 2 is zero

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We can solve the problem by using the law of conservation of momentum: the total momentum of the system must be conserved before and after the collision.

Therefore, we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 2000 kg is the mass of the first car

u_1 is the initial velocity of the first car

v_1 = 10 m/s is the final velocity of the first car (taking right as positive direction)

m_2 = 2000 kg is the mass of the second car

u_2 = 0 is the initial velocity of the second car

v_2 is the final velocity of the second car

We also know the initial momentum of car 1, which is

p_1 =40,000 kg m/s

And since momentum is the mass times the velocity, we find the initial velocity of car 1:

u_1 = \frac{p_1}{m_1}=\frac{40,000}{2,000}=20 m/s

with a positive sign, since the direction is to the right.

Now we can re-arrange the previous equation and solve for v2, the final velocity of car 2:

v_2 = \frac{m_1 u_1 -m_1 v_1}{m_2} = \frac{(2000)(20)-(2000)(10)}{2000}=10 m/s

And since the sign is positive, the direction is the same as the initial direction of car 1, so to the right.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

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