Initially, the velocity vector is 
. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 
, so the velocity is 
.
Converting back to direction and magnitude, we get 
Answer:
S=48.29 m
Explanation:
Given that the height of the hill h = 2.9 m
Coefficient of kinetic friction between his sled and the snow μ = 0.08
Let u be the speed of the skier at the bottom of the hill.
By applying conservation of energy at the top and bottom of the inclined plane we get.
Potential Energy=kinetic Energy
mgh = (1/2) mu²
u² = 2gh
u²=2(9.81)(2.9)
=56.89
u=7.54 m/s
a = - f / m
a = - μ*m*g / m
a = - μg
From equation of motion
v²- u² = 2 -μ g S
v=0 m/s
-(7.54)²=-2(0.06)(9.81)S
S=48.29 m
300 miles / 6 hours = 50 miles per hour
A wooden log is displaced to a distance of 20m in 10 seconds by applying 500N effort . Calculate the workdone and power...
Solution,
displacement = 20 m
time = 10 sec
force = 500 N
work done = ?
power = ?
Now ,
work done = f × s
= 500 N × 20 m
= 10000 j
Now ,
~nightmare 5474~
The input work = Force x distance
So I/P = 45 x 1.5 = 67.5 N.m
The output work = Force x distance
So, O/P = 87.6 x 0.4 = 35.04 N.m
Efficiency = (output work / Inputwork) x 100
= ( 35.04/67.5) x100
Efficiency = 51.91 %
