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pentagon [3]
2 years ago
7

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

think the coefficient of friction between the sled and the snow is. You've been walking at a steady 1.5\;{\rm m}/{\rm s}, and the rope pulls up on the sled at a 42.0^\circ angle. You estimate that the mass of the sled, with your friend on it, is 73.0 kg and that you're pulling with a force of 87.0 N.?

Physics
1 answer:
Juli2301 [7.4K]2 years ago
5 0

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used  to solve this question.

From the diagram, the horizontal component of the force is:

F_X = F_{cos \ \theta}

Replacing 42°  for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

F_X =64.65 \ N

On the other hand, the vertical component  is ;

F_Y = Fsin \ \theta

Replacing 42°  for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

F_Y =58.21  \ N

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

f = \mu \ N

Where;

\mu is said to be the coefficient of the friction

N = the  normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing F_Y \ for \ F_{sin \  \theta}. The normal reaction can now be:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

F_X = \mu \ ( mg - \ F_Y)

Making \mu the subject of the formular in the above equation; we have the following:

\mu \ = \ \frac{F_X}{mg - F_Y}

Replacing the following values: i.e

F_X \ = \ 64.65 \  N

m = 73 Kh

g  = 9.8 m/s²

F_Y = \ 58.21 N

Then:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

\mu = 0.0984

Thus, the coefficient of friction is = 0.0984

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Explanation:

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2 years ago
What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal
murzikaleks [220]
Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

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Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of v_f = 16.5 m/s, so \Delta v=16.5 m/s, in a time interval of \Delta t = 181 ms=0.181 s. The acceleration of the ball in the horizontal direction (x-axis) is therefore

a_x =  \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2

And the distance covered by the ball during this time interval, before it is released, is:

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(b) For this part we need to consider also the weight of the ball, which is W=mg=2.28 N

From this, we find its mass: m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg

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F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

F_y=W=mg=2.28 N (upward)

So, the magnitude of the force is

F= \sqrt{F_x^2+F_y^2}=  \sqrt{(20.98N)^2+(2.28N)^2}=21.2 N

To find the direction, we should find the angle of F with respect to the horizontal. This is given by

\tan \alpha =  \frac{F_y}{F_x}= \frac{2.28 N}{20.98 N}=0.11

From which we find \alpha=6.2^{\circ}

7 0
2 years ago
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SCORPION-xisa [38]

Answer:

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Explanation:

From the question we are told that

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Answer: d

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Answer:

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