A car moving in a straight line starts at x=0 at t=0. it passes the point x=25.0m with a speed of 11.0m/s at t=3.0s. it passesth
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Answer:
The horizontal distance is 4.823 m
Solution:
As per the question:
Mass of man, m = 65.0 kg
Height of the hill, H = 5.00 m
Mass of the backpack, m' = 20.0 kg
Height of ledge, h = 2 m
Now,
To calculate the horizontal distance from the edge of the ledge:
Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.
Now,
where
KE = Kinetic energy
PE = Potential energy
Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.
Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.
Therefore,
where
v = velocity at the hill's bottom
Now,
Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:
Now, time taken for the fall:
Now, the horizontal distance is given by:
x = v't =
Answer:
4 m/
Explanation:
From Equilibrium of forces, The Tension in string is cancelled by the Weight (product of mass and acceleration due to gravity) of the body acting downwards.
The Net force = Mass * Acceleration.
Since Net Force = 20 Newton, Mass = 5kg, therefore;
20 = 5kg * acceleration. Dividing the RHS and LHS of the equation by 5, we have;
Acceleration = which gives 4.
Note: RHS means Right Hand Side.
LHS means Left Hand Side.