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hoa [83]
3 years ago
15

A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 100 N is being applied to the sled rope

parallel to the ground, what is the force of friction between the sled and the snow?
Physics
1 answer:
tangare [24]3 years ago
3 0

Answer:

60.18 N

Explanation:

Given that:

The force applied on the sled = 100 N

Suppose, the angle between the sled rope and the ground = 53°

The horizontal force which acts  in the horizontal direction can be expressed as:

F_x = F \ cos \theta

F_x = 100 \ cos (53)

F_x = 60.18 \ N

But if the angle between the sled rope is parallel to the ground. Then, we use an angle on a straight line which is = 180°

F_x = F \ cos \theta

F_x = 100 \ cos (180)

= 100 × -1

= -100 N

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To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r
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Answer:

Toyota Camry

     F_d = 51.852 N

     F_d = 100.042 N

Hummer H2

     F_d = 412.0888 N

     F_d = 8351.755 N

Explanation:

Given:

- The density of air p_air = 1.2 kg/m^3

- The drag force equals car's engine force.

Find:

- What are the drag forces in newtons at 80 km/h and 105 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.)

- What are the drag forces in newtons at 80 km/h and at 105 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.)

Solution:

- The formula for drag force is given as follows:

                                 F_d = 0.5*C_d*p_air*A*V^2

Where,

A : The drag Area  m^2

C_d: The drag coefficient

V: Velocity  m/s

a)  Toyota Camry

                  C_d = 0.28 , A = 0.70 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.28*1.2*0.70*(22.22)^2

                  F_d = 51.852 N

                 C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.28*1.2*0.70*(29.1667)^2

                  F_d = 100.042 N

b)   Hummer H2

                 C_d = 0.57 , A = 2.44 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.57*1.2*2.44*(22.22)^2

                  F_d = 412.0888 N

                 C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.57*1.2*2.44*(29.1667)^2

                  F_d = 8351.755 N

4 0
3 years ago
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