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1 year ago

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A 161 lb block travels down a 30° inclined plane with initial velocity of 10 ft/s. If the coefficient of friction is 0.2, the to

tal work done, in ft-lbs by all forces when the block moves through a distance of 5 feet most nearly is?

1 answer:

lbvjy [14]1 year ago

5 0

Answer:

8418 ft lbf

Explanation:

Let g = 32 ft/s2. The gravitational force that is parallel to the incline is

$mgsin30^o = 161* 32* 0.5 = 2576 lbf$

The normal force that acting on the block is the gravity force component that is perpendicular to the incline:

$N = mgcos30^o = 32*161*\sqrt{3}/2 \approx = 4461.8 lbf$

The friction force is the product of normal force and coefficient

$F_f = \mu N = 0.2*4461.8 = 892.4 lbf$

So the net force is force of gravity subtracted by friction force

F = 2576 - 892.4 = 1683.6 lbf

So the work by this force over 5 feet distance is

W = Fs = 1683.6*5 = 8418 ft lbf

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A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

$s=1.63 m$

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1 year ago

Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha

aliina [53]

Answer:

P=627.47W

Explanation:

To solve this problem we have to take into account, that the work done by the winch is

$W=Fh$

the force, at least must equal the gravitational force

$F=Mg=(156kg)(9.8\frac{m}{s^2})=1258.8N$

with force the tension in the cable makes the winch go up.

The work done is

$W=(1258.8N)(58.0m)=73010.4J$

To calculate the power we need to know what is the time t. But first we have to compute the acceleration

The acceleration will be

$v_f^2=v_0+2ah\\a=\frac{v_f^2}{2h}=\frac{(24.9\frac{m}{s})}{2(58.0m)}=0.214\frac{m}{s^2}$

and the time t

$v_f=v_0+at\\t=\frac{v_f}{a}=116.35s$

The power will be

$P=\frac{W}{t}=\frac{73010.4J}{116.35s}=627.47W$

HOPE THIS HELPS!!

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2 years ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

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2 years ago

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emmasim [6.3K]

Answer:

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Explanation:

Please see the attachment below for film solution.

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1 year ago

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pochemuha

Answer:

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