There\'s a few questions under here------------ A 4.00 kg block located on a hor
ID: 1415024 • Letter: T
Question
There's a few questions under here------------
A 4.00 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.0 N at an angle = 35.0° above the horizontal, as shown in the Figure.
https://s4.lite.msu.edu/enc/84/e43a84d123cd29b2cd40787b8d310ae5c264fec7d316b74e2edee43f267afdca54ed126ca22ee270a671d56b41ea6f21fa47d07ec5726fbf57934eae302c76c2d091f86094278cb296bda564333e18a29a3e7fae437e3cb2d8c658db90c45c25215b7267679e6a8a8d6031be0a46af80.gif
The coefficient of kinetic friction between the block and the floor is 0.100. What is the speed of the block 5.10 s after it starts moving?
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A hockey puck on a frozen pond with an initial speed of 14.7 m/s stops after sliding a distance of 197.1 m. Calculate the average value of the coefficient of kinetic friction between the puck and the ice.
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Mass m1 = 13.3 kg is on a horizontal surface. Mass m2 = 8.13 kg hangs freely on a rope which is attached to the first mass. The coefficient of static friction between m1 and the horizontal surface is s = 0.622, while the coefficient of kinetic friction is k = 0.115.
https://s4.lite.msu.edu/enc/54/e43a84d123cd29b211bedeb8f2fce09234b7f9794e34d271bf0738e51b036329680216764a60e180e7963fa80b895182a8a9f3efac8cfc093710352375df741228d9bf34826ceeba.gif
If the system is set in motion with m1 moving to the right, then what will be the magnitude of the system's acceleration? Consider the pulley to be massless and frictionless.
If the system is set in motion with m1 moving to the left, then what will be the magnitude of the system's acceleration?
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The three blocks shown are relased at t=0 from the position shown in the figure. Assume that there is no friction between the table and M2, and that the two pulleys are massless and frictionless. The masses are: M1 = 3.0 kg, M2 = 9.0 kg, M3 = 5.0 kg.
https://s4.lite.msu.edu/enc/84/c19eb57f8b48197543ce2c632dae5ee18bc32e2000317f8cfaa313edcd1ed2b1024ed589b8d11b46ec1dfb34cc31a953147da16ddb86760441a42f2c52af377f0184c1a7f2dbc46088c19b2255aa84cbd4bbf763a00156c68a3fddbb162b9bb4fa38264c4493b8db2da0715ff2e3ef8f.gif
Calculate the speed of M2 at a time 1.35 s after the system is released from rest.
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The initial velocity of a 1.45 kg block sliding down a frictionless inclined plane is found to be 1.29 m/s. Then 1.08 s later, it has a velocity of 5.50 m/s.
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What is the angle of the plane with respect to the horizontal?
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A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 34.5°, the block starts to slide down the incline, traveling 3.50 m down the incline in 2.40 s. Calculate the coefficient of static friction between the block and the plank.
Calculate the kinetic coefficent of friction between the block and the plank.
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A block with mass m = 12.4 kg slides down an inclined plane of slope angle 24.1o with a constant velocity. It is then projected up the same plane with an initial speed 1.50 m/s. How far up the incline will the block move before coming to rest?
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Explanation / Answer
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A hockey puck on a frozen pond with an initial speed of 14.7 m/s stops after sliding a distance of 197.1 m. Calculate the average value of the coefficient of kinetic friction between the puck and the ice.
vi = 14.7 m/s
s = 197.1 m
Friction Force, Ff = uk*m*g
m*a = uk*m*g
a = uk*g
vf^2 = vi^2 + 2*a*s
0 = 14.7^2 - 2*a*197.1
a = 14.7^2/(2*197.1)
uk * 9.8 = 14.7^2/(2*197.1)
uk = 0.056
A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 34.5°, the block starts to slide down the incline, traveling 3.50 m down the incline in 2.40 s. Calculate the coefficient of static friction between the block and the plank.
Let coefficient of static friction between the block and the plank, be us
Friction Force, Ff = us*m*g*cos(34.5)
us*m*g*cos(34.5) = m*g*sin(34.5)
us = tan(34.5)
us = 0.687
Let kinetic coefficent of friction between the block and the plank be uk
s = 3.5 m
t = 2.4 s
s = u*t + 1/2*a*t^2
3.5 = 0 + 1/2 * a * 2.4^2
a = 1.215 m/s^2
Now,
m*g*sin(34.5) - uk*m*g*cos(34.5) = m*a
9.8*sin(34.5) - uk*9.8*cos(34.5) = 1.215
uk = 0.536
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