Which is the hottest and coldest planet in solar system ?
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The force of friction is <u>34.3 N.</u>
A block of mass m slides down a plane inclined at an angle θ to the horizontal with a constant velocity. According to Newton's first law of motion, every body continues in its state of rest or a state of uniform motion in a straight line, unless acted upon, by an external unbalanced force. This means that when balanced forces act on a body, the body moves with a constant velocity.
The free body diagram of the sliding block is shown in the attached diagram. Resolve the weight mg of the block into two components mg sinθ along the direction of the plane and mg cosθ perpendicular to the plane . The force of friction F acts upwards along the plane and the normal reaction acts perpendicular to the plane.
Since the block moves down with a constant velocity, the downward force mg sinθ must be equal to the upward frictional force.
Substitute 7 kg for m, 9.8 m/s² for g and 30° for θ.
The force of friction is <u>34.3 N</u> up the plane.
Answer:
the coin does not slide off
Explanation:
mass (m) = 5 g = 0.005 kg
distance (r) = 15 cm = 0.15 m
static coefficient of friction (μs) = 0.8
kinetic coefficient of friction (μk) = 0.5
speed (f) = 60 rpm
acceleration due to gravity (g) = 9.8 m/s^{2}
lets first find the angular speed of the table
ω = 2πf
ω = 2 x π x 60 x
ω = 6.3 s^{-1]
Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding
static force (Fs) = ma
static force (Fs) = μs x Fn = μs x m x g
Fs = 0.8 x 0.005 x 9.8 = 0.0392 N
Fs = ma
0.0392 = 0.005 x a
a = 7.84 m/s^{2}
= a x r
= 7.84 x 0.15
Vmax = 1.08 m/s
ωmax =
ωmax = = 7.2 s^{-1}
now that we have the maximum angular acceleration of the table, we can calculate its maximum speed in rpm
Fmax =
Fmax = = 68.7 rpm
since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.