Answer:
(a) Magnitude of static friction force is 109 N
(b) Minimum possible value of static friction is 0.356
Solution:
As per the question;
Horizontal force exerted by the girl, F = 109 N
Mass of the crate, m = 31.2 kg
Now,
(a) To calculate the magnitude of static friction force:
Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:
F = f = 109 N
(b) The maximum possible force of friction between the floor and the crate is given by:
where
N = Normal reaction = mg
Thus
For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.
Australia separated from other continents and species there evolved independently
Answer:
(a) 1.093 rad/s^2
(b) 4.375 rad/s
(c) 8.744 rad/s
(d) 67.845 rad
Explanation:
initial angular velocity, ωo = 0
time, t = 8s
angular displacement, θ = 35 rad
(a) Let α be the angular acceleration.
Use second equation of motion for rotational motion
By substituting the values
35 = 0 + 0.5 x α x 8 x 8
α = 1.093 rad/s^2
(b) The average angular velocity is defined as the ratio of total angular displacement to the total time taken .
Average angular velocity = 35 / 8 = 4.375 rad/s
(c) Let ω be the instantaneous angular velocity at t = 8 s
Use first equation of motion for rotational motion
ω = ωo + αt
ω = 0 + 1.093 x 8 = 8.744 rad/s
(d) Let in next 5 seconds the angular displacement is θ.
By substituting the values
θ = 8.744 x 5 + 0.5 x 1.093 x 5 x 5
θ = 67.845 rad
