The first thing you should know for this case is that work is defined as the product of force by the distance traveled in the direction of force.
We have then:
W = Fd
The distance varies, so we must integrate:
from 0 to 20:
W = ∫F (x) dx
W = ∫32xdx
W = 32∫xdx
W = 32 (x ^ 2/2) = (16) (20 ^ 2) = 6400 ft * lbs
answer:
6400 ft * lbs is work done pulling the rope up 20 ft
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.
Answer:
Speed of lighter ball is 4 m/s.
Explanation:
Applying the principle of conservation of linear momentum,
momentum before collision = momentum after collision.

+
= 
- 

= 3 kg,
= 8 m/s,
= 2 kg,
= 0 m/s ( since it is at rest),
= 2 m/s,
= ?
(3 x 8) + (2 x 0) = (8 x 2) - (2 x
)
24 + 0 = 16 - 2
2
= 16 - 24
2
= -8
= 
= -4 m/s
This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.
Answer:
The ball will reach the ground in 0.8s
Option C
Explanation:
Given:
- Takes t = 0.8 s for ball to reach ground when thrown horizontal from top of a building.
Find:
If it had been thrown with twice the speed in the same direction, it would have hit the ground in how many second.
Solution:
- We know that the amount of time taken to hit the ground is determined by the vertical distance i.e height at which it is thrown. The displacement of ball from top is given by:
S_y = S_o + V_i,y*t + 0.5*g*t^2
- We know that the S_o = height of the building.
We also know that the ball os thrown horizontally; hence, y-component of initial velocity is zero. V_y,i = 0
0 = h + 0 + 0.5*g*t^2
- Hence, the time taken t is:
t = sqrt ( 2h / g)
- The time taken to reach the ground is independent of the initial speed. Hence, the ball will reach the ground in 0.8s .