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3 years ago

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

1 answer:

3 0

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
X: mgsin65 - F = mAx 
Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) 
Moment about center of mass: 
Fr = Iα 
Now Ax = rα 
and F = umgcos65 
mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) 
umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) 
ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) 
ugcos65 = 0.4(gsin65 - ugcos65) 
1.4ugcos65 = 0.4gsin65 
1.4ucos65 = 0.4sin65 
u = 0.4sin65/1.4cos65 
u = 0.613

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3 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

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3 years ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

brilliants [131]

Answer:

Explanation:

When the craft was stationary , weight will be balanced by tension

T = mg

T = 7000 N

when the craft was being lowered to the seafloor

drag force will act in upper direction , so

T₁ + 1800 = mg

T₁ = mg - 1800

= 7000 - 1800

= 5200 N

52 X 10² N

when the craft was being raised from the seafloor , Tension will act in downward direction

T₂ = mg+ 1800

T₂ = 7000 - 1800

= 8800N

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