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Nostrana [21]
3 years ago
5

An archer pulls back the string of a bow to release an arrow at a target. Which kind of potential energy is transformed to cause

the motion of the arrow?
Physics
1 answer:
erastovalidia [21]3 years ago
7 0
I believe the answer to this would be elastic.
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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
How fast can Usain Bolt run if it takes him 9.9 s to run 100m?​
Arte-miy333 [17]

Answer:

27.8 mph

Explanation:

May I have brainliest please? :)

8 0
3 years ago
Read 2 more answers
A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it
Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

L = length of the meter stick = 1 m

m = mass of the meter stick

w = angular speed of the meter stick as it hits the floor

v = speed of the other end of the stick

we know that, linear speed and angular speed are related as

v = r w\\w = \frac{v}{r}

h = height of center of mass of meter stick above the floor = \frac{L}{2} = \frac{1}{2} = 0.5 m

I = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

I = \frac{mL^{2} }{3}

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}

7 0
3 years ago
Why do the waves have different speeds in different layers of Earth's surface?
Genrish500 [490]

Answer:

Material's density

Explanation:

Seismic waves travel at different rates of speed based on a material's density. Hopefully, you understand that the Earth has three main layers: the crust, mantle, and core. Earthquake waves move faster through solids.

6 0
1 year ago
Should i beat up my sibling cause they mean to me?
Vlada [557]

no thats messed up lol

4 0
3 years ago
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