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REY [17]
3 years ago
11

A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting

motion, including friction and air resistance, total 450 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude (in N) of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?
Physics
1 answer:
nata0808 [166]3 years ago
6 0

Answer:

1185 N

Explanation:

From Newton’s second law of motion,  

F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force

Net force acting on motorcycle F_{net} is given by  F_{net}=F-f

Where F is force acting on motorcycle and f is frictional force

Substituting F-f for F_{net}

F_{net}=ma hence  ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided

245*3= F- 450

F=245*3 +450= 1185 N

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Answer:

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Explanation:

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    sin52 = Wx / W

    cos52 = Wy / W

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Let's write them equations

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    Wx = ma

Y Axis

    N-Wy = 0

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a) Let's calculate the acceleration

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Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

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d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

F - External force exerted on the sled, measured in newtons.

f - Friction force, measured in newtons.

N - Normal force from the ground on the mass, measured in newtons.

W - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

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W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

W = 490.35\,N

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

F_{min,s} = \mu_{s}\cdot W (2)

Where:

\mu_{s} - Static coefficient of friction, dimensionless.

W - Weight of the sled, measured in newtons.

If we know that \mu_{s} = 0.3 and W = 490.35\,N, then the force needed to start the sled moving is:

F_{min,s} = 0.3\cdot (490.35\,N)

F_{min,s} = 147.105\,N

A force of 147.105 newtons is needed to start the sled moving.

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F_{min,k} = \mu_{k}\cdot W (3)

Where \mu_{k} is the kinetic coefficient of friction, dimensionless.

If we know that \mu_{k} = 0.1 and W = 490.35\,N, then the force needed to keep the sled moving at a constant velocity is:

F_{min,k} = 0.1\cdot (490.35\,N)

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