from the figure
let the total distance is x
x = vi*t+1/2*a*t^2
here acceleration = a
from the figure
ma = mgsin40-
mgcos40
a = 9.8sin40-
*9.8cos40
= 6.301-
7.51
x = 1/2(6.301-
7.51)*10^2
= 315.05-
*375.5
at the 5 sec
speed vf = ( 6.301-
7.51)*5
= 31.505-
*37.55
for last 5 sec
200 = (31.505-
*37.55)*5+1/2*(6.301-
7.51)*5^2
200 = 236.2875-
*281.625
= 0.129 (answer for part B)
length of the ramp
x = 315.05-0.129*375.5 = 266.61 m (answer for part A)
Problem 3 A block accelerates at a constant rate down a long inclined ramp. The coefficient...
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A 15kg block slides up a 30∘ inclined plane at a constant velocity. The coefficient of friction between the block and the plane is μk=0.2. In order for the block to slide up the incline, someone must apply a force to the block. What is the magnitude of the applied force if it points parallel to the incline? What is the magnitude of the applied force if it points horizontally instead?
A 15kg block slides up a 30∘ inclined plane at a constant velocity. The coefficient of friction between the block and the plane is μk=0.2. In order for the block to slide up the incline, someone must apply a force to the block. Part A: What is the magnitude of the applied force if it points parallel to the incline? Part B: What is the magnitude of the applied force if it points horizontally instead?
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A block of mass m = 3.5 kg is
on an inclined plane with a coefficient of friction
μ1 = 0.31, at an
initial height h = 0.53 m above
the ground. The plane is inclined at an angle θ =
54°. The block is then compressed against
a spring a distance Δx = 0.11 m
from its equilibrium point (the spring has a spring constant of
k1 = 39 N/m) and
released. At the bottom of the inclined plane...
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