relation between linear velocity and angular velocity is given as
here
v = linear speed
R = radius
= angular speed
now plug in all data in the equation
so rotating speed is 60.9 rad/s
You want to use a rope to pull a 10-kg box of books up a plane inclined 30∘ above the horizontal. The coefficient of kinetic fri
ction is 0.29. The rope pulls parallel to the incline.
What force do you need to exert on the rope if you want to pull the box with a constant acceleration of 0.50 m/s2 up the plane?
What force do you need to exert on the rope if you want to pull the box with a constant acceleration of 0.50 m/s2 up the plane?
1 answer:
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Answer:
(i) 12 seconds
(ii) 216 meters from the initial position
(iii) 132 meters from the initial position
(iv) No
Explanation:
Speed of express train =36 m/s
Speed of local train =11 m/s
The initial distance between the local train and passenger train =100 m.
Due to the application of breaks, the express train slows at the rare of
So, the acceleration of the express train, .
(i) Let t be the time the express train takes to stop.
From the equation of motion,
v=u+at
where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.
In this case, v=0, u=36 m/s,
So, 0=36+(-3)t
seconds.
(ii) To compute the distance traveled, s, till the express train stops, using
meters.
(iii) The local train is moving at a speed of 11 m/s
So, in 12 seconds, the distance, d, traveled by the local train
d= 11x12=132 meters [as distance= speed x time]
(iv) Let 0 be the reference position which is the initial position of the express train.
So, at the initial time, the position of the local train is at 100m.
After 12 seconds:
The position of the express train is at 216 m [using part (ii)]
and the position of the local train is at 100+132=232m [using part (iii)].
So, the local train is still ahead of the express train, hence the trains didn't collide.