Potential energy
potential energy is the stored energy an object has, the bike is not moving and it just sitting on the hill, with the stored energy
The force of gravity between two objects is:
F = G*m1*m2/r^2
So, it is dependent of the two masses and the distance between their centers of mass.
Answer:
0.20
Explanation:
The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.
There are two forces acting in the horizontal direction:
- The pushing force: F = 99 N, forward
- The frictional force:
, backward, with
coefficient of kinetic friction
m = 50 kg mass of the box
g = 9.8 m/s^2 gravitational acceleration
The net force must be zero, so we have

which we can solve to find the coefficient of kinetic friction:

'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.
For the stone dropped from the 50m tower:
H = +50 - (1/2) G T²
For the stone tossed upward from the ground:
H = +20T - (1/2) G T²
When the stones' paths cross, their <em>H</em>eights are equal.
50 - (1/2) G T² = 20T - (1/2) G T²
Wow ! Look at that ! Add (1/2) G T² to each side of that equation,
and all we have left is:
50 = 20T Isn't that incredible ? ! ?
Divide each side by 20 :
<u>2.5 = T</u>
The stones meet in the air 2.5 seconds after the drop/toss.
I want to see something:
What is their height, and what is the tossed stone doing, when they meet ?
Their height is +50 - (1/2) G T² = 19.375 meters
The speed of the tossed stone is +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it. The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.
Answer:
A.) 1372 N
B.) 1316 N
C.) 1428 N
Explanation:
Given that a 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following cases:
a. The load moves downward at a constant velocity
At constant velocity, acceleration = 0
T - mg = ma
T - mg = 0
T = mg
T = 140 × 9.8
T = 1372N
b. The load accelerates downward at a rate 0.4 m/s??
Mg - T = ma
140 × 9.8 - T = 140 × 0.4
1372 - T = 56
-T = 56 - 1372
- T = - 1316
T = 1316N
C. The load accelerates upward at a rate 0.4 m/s??
T - mg = ma
T - 140 × 9.8 = 140 × 0.4
T - 1372 = 56
T = 56 + 1372
T = 1428N