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solmaris [256]
3 years ago
7

Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo

r. The tension is a constant 30 N and the coefficient of kinetic friction is 0.20. Use work and energy to find Susan’s speed after being pulled 3.0 m.

Physics
2 answers:
kifflom [539]3 years ago
5 0

Answer: speed= 1.95866m/s ~= 2.0 m/s

Explanation: check attachment for explanation

kiruha [24]3 years ago
4 0

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

W=F*d*cos(\theta)

here we have the work done by Paul and the friction force, so:

W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J

W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J

Now the change of energy is:

W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s

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laila [671]

Answer:

The ball is dropped at a height of 9.71 m above the top of the window.

Explanation:

<u>Given:</u>

  • Height of the window=1.5 m
  • Time taken by ball to cover the window height=0.15

Now using equation of motion in one dimension we have

s=ut+\dfrac{at^2}{2}

Let u be the velocity of the ball when it reaches the top of the window

then

1.5=0.15u+\dfrac{9.8\times0.15^2}{2}\\u=3.96\ \rm m/s\\

Now u is the final velocity of the ball with respect to the top of the building

so let t be the time taken for it to reach the top of the window with this velocity

3.96=gt\\t=0.4\ \rm s\\

Let h be the height above the top of the window

h=\dfrac{gt^2}{2}\\\\h=\dfrac{9.8\times 0.4^2}{2}\\h=9.71\ \rm m

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3 years ago
Which scenario involves kinetic energy transforming into potential energy? A. a book lying on a shelf B. a train traveling at a
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The answer is C.

The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
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3 years ago
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8 0
3 years ago
Read 2 more answers
Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light
SIZIF [17.4K]

Answer:

The correct answer is "4.443 sec".

Explanation:

Given:

Mass of child,

= 34 kg

Mass of swing,

= 18 kg

Length,

= 4.9 m

The time period of pendulum will be:

T = 2 \pi \sqrt{4g}

  = 2 \pi \sqrt{\frac{4.9}{9.8} }

  = 4.443 \ sec  

5 0
3 years ago
Read 2 more answers
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
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