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3 years ago

11

A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t

he pedals, how many joules of output work are used to move the bicycle?

1 answer:

emmainna [20.7K]3 years ago

8 0

Efficiency = Work Output / Work Input

92% = Work Output / 100

0.92 = Work Output / 100

Work Output = 0.92 * 100

Work Output = 92 joules.

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

rjkz [21]

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:

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

which is our answer to part a.

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.

3 0

3 years ago

Read 2 more answers

A ball rolls off a desk at a speed of 3 m/s and lands .40 seconds later. How far from the base of the desk does the ball land?

Salsk061 [2.6K]

Is the velocity constant? Is there any friction?

3 meters per second

then after 40 seconds it must 3*40 = 120 meters

120 meters or 0.12 km if you will

7 0

3 years ago

How is a family defined

natulia [17]

Answer:Family

Explanation: Family is defined as people you love, people that you can count on. People that make you happy. Family isnt only people who are related to you by blood, we are all a family. Certain people are your family.

3 0

3 years ago

A skier is accelerating down a 30.0-degree hill at 3.80 m/s^2.

Bond [772]

Answer:

ax = -3.29[m/s²]

ay = -1.9[m/s²]

Explanation:

We must remember that acceleration is a vector and therefore has magnitude and direction.

In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.

$a_{x}=-3.8*cos(30) = -3.29 [m/s^{2}]\\ a_{y}=-3.8*sin(30) = -1.9 [m/s^{2}]$

3 0

3 years ago

The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the

natta225 [31]

Answer:

50 N

Explanation:

Efficiency of a machine can't be more than 1, so I assume you mean 40%. (Remember, efficiency and mechanical advantage are not the same).

Efficiency is the ratio of work out of a system to the work in to the system.

e = Wout / Win

Work is force times distance, so:

e = (Fout × Dout) / (Fin × Din)

Rearranging:

Fin = (Fout × Dout) / (e × Din)

Fin = (Fout / e) × (Dout / Din)

Fin = (Fout / e) / (Din / Dout)

We know that e = 0.40, and Fout = 120 N. Since there are 6 pulleys, we also know that Din/Dout = 6.

F = (120 N / 0.4) / 6

F = 50 N

5 0

3 years ago