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kenny6666 [7]
2 years ago
10

A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient

of sliding friction between the sidewalk and the metal runners of the sled?

Physics
1 answer:
Andre45 [30]2 years ago
5 0

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

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A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. T
irina1246 [14]

Answer:

Option d)

Solution:

As per the question:

Work done by farm hand, W_{FH} = 972J

Force exerted, F' = 310 N

Angle, \theta = 23^{\circ}

Now,

The component of force acting horizontally is F'cos\theta

Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.

Thus

W_{FH} = \vec{F'}.\vec{d}

972 = 310\times dcos23^{\circ}

d = 3.406 m = 3.4 m

3 0
2 years ago
Consider a boat heading due east at 15 miles/hour. The water's current is moving at 7.1 miles/hour at 45º south of east. Drag ve
givi [52]

If a boat is going East at 15mph and there is a water current going southeast at 45° then the boat is being drifted southward.  So since the current is going at an angle then it has a x and y component.  So Rx refers to the x-component force of the current and Ry refers to the y-component of the current, and |R| refers to the magnitude of these forces.

7 0
2 years ago
Read 2 more answers
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
2 years ago
The total energy of a 0.050 kg object travelling at 0.70 c is:
finlep [7]
What is the unit c denotes here
3 0
2 years ago
Read 2 more answers
Norma kicks a soccer ball with an initial velocity of 10.0 meters per second at an angle of 30.0°. If the ball moves through the
enyata [817]

<u>Answer</u>

27.7


<u>Explanation</u>

The ball was hit at an angle of 30°, with the horizontal at a speed of 10 m/s. We have to find the horizontal component of speed.

cosx = adjacent/hypotenuse

cos 30 = adjacent / 10

adjacent = 10 cos30

             = 8.66 m/s        ⇒ This is the horizontal speed.

Now  find the horizontal distance.

Distance = speed × time

               = 8.66 × 3.2

                = 27.71

Answer to the nearest tenth = 27.7

4 0
2 years ago
Read 2 more answers
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