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Kryger [21]
2 years ago
5

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?
Physics
1 answer:
lbvjy [14]2 years 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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iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0
2 years ago
Two hockey players skating on essentially frictionless ice collide head-on. Madeleine, of mass 65.0 kg, is moving at 6.00 m/s to
xeze [42]

Explanation:

It is given that,

Mass of Madeleine, m_1=65\ kg

Initial speed of Madeleine, u_1=6\ m/s (due east)

Final speed of Madeleine, v_1=-3\ m/s (due west)

Mass of Buffy, m_2=55\ kg

Final speed of Buffy, v_2=3.5\ m/s (due east)

Let u_1 is the Buffy's velocity just before the collision. Using the conservation of linear momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

65\times 6+55\times u_2=65\times (-3)+55\times 3.5

u_2=-7.13\ m/s

So, the initial speed of the Buffy just before the collision is 7.13 m/s and it is moving due west. Hence, this is the required solution.

5 0
2 years ago
A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?
GarryVolchara [31]
Answer: Δp = F*Δt = 500N*0.001s = 0.5Ns
3 0
2 years ago
Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th
Stels [109]

This question deals with the law of conservation of momentum, which basically says that the total momentum in a system must stay the same, provided there are no outside forces. Since you were given the mass and velocity of the two objects you can find the momentum (p=mv) of each and then add them together to find the total momentum of the system before they collide. This total momentum must be the same after they collide.  Since you have the mass and velocity of one of the objects after the collision you can find the its momentum after.  Subtract this from the the system total and you will have the momentum of the other object after the collision.  Now that you know the momentum of the other object you can find its velocity using p=mv and its mass from before.

Be careful with the velocities.  They are vectors, so direction matters.  Typically moving to the right is positive (+) and moving to the left is negative (-).  It is not clear from your question which direction the objects are moving before and after the collision.

6 0
2 years ago
Read 2 more answers
A transverse wave is traveling on a string stretched along the horizontal x-axis. The equation for the
maxonik [38]

Answer:

A) 0.33 m/s

Explanation:

The standard form of a transverse wave is given by  

y = a  cos ( ω t − kx ) ,   k =  2 π  / λ

Amplitude,   a =  0.002  m

Wavenumber (k)=47.12 and wavelength  ( λ )  =  0.133 m

Time period(T)=0.0385 s and angular frequency  ( ω )  =  52 π  rad/s

Maximum speed of the string is given by  aw

Therefore ; max. speed = 0.002 x 52 π = 0.327 m/s

 

6 0
2 years ago
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