Starting from rest at the top, a child slides down the water slide at a swimming pool and enters the water at a final speed of 4
1 answer:
Energy conservation law states:
At the top potential energy is at maximum and kinetic energy is zero. At the bottom potential energy is zero and kinetic energ is at maximum.
We have:
We can use this formula to calculate height:
To calculate requested speed we just need to double the height and insert it into equation:
At the top potential energy is at maximum and kinetic energy is zero. At the bottom potential energy is zero and kinetic energ is at maximum.
We have:
We can use this formula to calculate height:
To calculate requested speed we just need to double the height and insert it into equation:
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Answer:
kinetic friction may be greater than 400 N or smaller than 400 N
Explanation:
As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as
now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as
now we know that
so here value of limiting static friction force is always more than kinetic friction
also we know that
initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction
and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction
so kinetic friction may be greater than 400 N or smaller than 400 N
Answer:
a) 20 rad/s
b) 6 m/s
Explanation:
b) Force acting on the wheel is 200 N
mass of the wheel is 100 kg
From Newton's second law of motion, F = m × a
Where F is the net force acting on the body
m is mass of the body
a is the acceleration of the body
By substituting the values we get, a = 2 m/s²
As acceleration is constant, we can use the below formula for calculating the final velocity of the object
v = u + a × t
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
u = 0 (∵ it starts from rest)
By substituting the values we get
v = 0 + 2 × 3 = 6 m/s
∴ Speed of center of mass after 3 seconds = 6 m/s
a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel
∴ Radius of the wheel = 300 mm
Torque acting on the wheel about axis of rotation = 300 mm × 200 N =
60 N·m
Torque = (Moment of inertia) × (angular acceleration)
Assuming that the mass of spokes of the wheel to be negligible,
Moment of inertia of the wheel about axis of rotation = 100 × 300² × = 9 kg·m²
Then,
60 = 9 × (angular acceleration)
∴ angular acceleration ≈ 6·67 rad/s²
As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed
=
+ α × t
Where
is the final angular velocity
is the initial angular velocity
α is the angular acceleration
t is the time taken
is 0 (∵ initially it starts from rest)
By substituting the values we get
= 6·67 × 3 = 20 rad/s
∴ Angular velocity of the wheel after three seconds = 20 rad/s