Answer:
Explanation:
Given data
Mass m=120.0 kg
Coefficient of static friction
The coefficient of sliding friction
For Part (a) Maximum force
According to Newtons second law the net force aced on the body is given by
The friction force is given by
Conclude that
For Part (b) Acceleration
The acceleration due to dynamic friction is given by:
The dynamic friction is given by:
So the acceleration given by
Answer:
During its trajectory the cannon ball will have kinetic and potential energy
E = KE + PE
The maximum value of PE will occur when the cannon ball was fired straight up - for other trajectories PE will be less because of translational KE
So E = M g h = 1/2 M v^2 for maximum height
h = v^2 / (2 * g) = 400 / (2 * 9.8) = 20.4 m
675 nM is 675 x 10⁻⁹ meter ... 0.000 000 675 meter .
That's 0.000 067 5 of one centimeter
0.000 675 of one millimeter
Net force = (mass) x (acceleration)
There are two forces acting on the barrel:
-- the force of gravity, downward;
-- the force of the rope, upward.
When these two forces are added together, their sum (the net force
on the barrel) causes the barrel to accelerate upward at 1.4 m/s² .
Net force = (Rope force) - (Grav force) = (mass) x (acceleration)
Grav force = 'weight' = (mass) x (grav) = (16 x 9.8) = 156.8 newtons
Net force = (mass x acceleration) = (16 x 1.4) = 22.4 newtons
Net force =
22.4 newtons = (Rope force) - (156.8 newtons)
Add 156.8 newtons to each side: Rope force = 22.4 + 156.8 = <em>179.2 newtons
</em>