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Travka [436]
3 years ago
15

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

y other forces, what is the maximum acceleration a this engine can produce in 1650-kg car?
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

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Answer:

a. Technician A

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3 years ago
A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the
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Answer:

a) F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

b) \mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

\Sigma F_{y}=0

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

-F_{y}-W+N=0

When solving for the normal force we get:

N=F_{y}+W

we know that

W=mg

and

F_{y}=Fsin \theta

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

f_{k}=\mu_{k}N

so we can find the kinetic friction by substituting for N, so we get:

f_{k}=\mu_{k}(F sin \theta +mg)

Now we can find the sum of forces in x:

\Sigma F_{x}=0

so after analyzing the diagram we can build our sum of forces to be:

-f+F_{x}=0

we know that:

F_{x}=Fcos \theta

so we can substitute the equations we already have in the sum of forces on x so we get:

-\mu_{k}(F sin \theta +mg)+Fcos \theta=0

so now we can solve for the force, we start by distributing \mu_{k} so we get:

-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0

we add \mu_{k}mg to both sides so we get:

-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg

Nos we factor F so we get:

F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg

and now we divide both sides of the equation into (cos \theta-\mu_{k} sin \theta) so we get:

F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}

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Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

f_{s}=\mu_{s}(F sin \theta +mg)

and

F cos θ = f

when substituting one into the other we get:

F cos \theta=\mu_{s}(F sin \theta +mg)

which can be solved for the static friction coefficient so we get:

\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}

which is the answer to part b.

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