a man is 9 m behind a dorr of train when it starts moving with a=2ms^-1. how far the man have to run and after what time will he
1 answer:
You might be interested in
Explanation:
Given parameters:
Distance = 15miles north = 24140.2m
Initial velocity = 0m/s
Final velocity = 4m/s
Unknown:
Speed, velocity and acceleration = ?
Solution:
The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.
Speed =
The speed of the student is 4m/s
Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;
Velocity =
The velocity of the student is 4m/s due north
Acceleration is the change in velocity with time;
To find the acceleration, we use
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
4² = 0² + 2x a x 24140.2
a = = 0.00033m/s²
Answer:
The tension in the rope is 229.37 N.
Explanation:
Given:
Mass of the block is,
Coefficient of static friction is,
Angle of inclination is, °
Draw a free body diagram of the block.
From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.
Forces acting are and normal
. Now, there is no motion in the direction perpendicular to the inclined plane. So,
Consider the direction along the inclined plane.
The forces acting along the plane are and frictional force,
, down the plane and tension,
, up the plane.
Now, as the block is at rest, so net force along the plane is also zero.
Therefore, the tension in the rope is 229.37 N.
