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levacccp [35]
3 years ago
7

A man weighing 180 lbf pushes a block weighing 100 lbf along a horizontal plane. the dynamic coefficient of friction between the

block and plane is 0.2. assuming that the block moves at a constant speed, how much work is required to move the block 100 ft considering the (a) the block and (b) the man as the system.
Physics
1 answer:
Talja [164]3 years ago
3 0
The first thing you should know is that the work is defined as:
 W = F * d
 Where
 F = force
 d = displacement
 We have then
 (a) the block
 F = (0.2) * (100) = 20
 d = 100
 W = (20) * (100) = 2000 ft.lbf
 (b) the man as the system.
 F = (0.2) * (100 + 180) = 56
 d = 100
 W = (56) * (100) = 5600 ft.lbf
 answer:
 (a) 2000 ft.lbf
 (b) 5600 ft.lbf
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This equation can be differentiated with respect to the radius of change, that is

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At the same time since Newton's second law we know that:

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From the previous value given for acceleration we have to

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dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

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3 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

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Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

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Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

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