Kinetic energy and potential energy pair is the quantity in which one will increase then other will decrease
As we know that sum of kinetic energy and potential energy will always remain conserved
So here we will have
so here as we move away from mean position the kinetic energy will decrease while at the same time potential energy will increase.
So the pair of potential energy and kinetic energy will satisfy the above condition
Answer:
Option ( B ) is correct .
Explanation:
To lift a heavy weight , inclined plane is used . Use of inclined plane , makes the task easier because instead of force mg , force mg sinθ is to be used which is less than mg . Here θ is inclination of inclined plane.
If h be the height by which weight is to be lifted
potential energy acquired by weight = mgh
work done by force mg sinθ = mgsinθ x d where d is displacement required .
mg sinθ x d = mgh ( work done by force = potential energy stored in luggage )
d = h / sinθ
d will be more than h
Hence inclined plane increases the distance to be covered by force applied though it decreases the force itself.
Hence option ( B ) is correct .
Answer:
Kf= 36 J
W(net) = 32 J
Explanation:
Given that
m = 2 kg
F= 4 N
t= 2 s
Initial velocity ,u= 2 m/s
We know that rate of change of linear momentum is called force.
F= dP/dt
F.t = ΔP
ΔP = Pf - Pi
ΔP = m v - m u
v= Final velocity
By putting the values
4 x 2 = 2 ( v - 2)
8 = 2 ( v - 2)
4 = v - 2
v= 6 m/s
The final kinetic energy Kf
Kf= 1/2 m v²
Kf= 0.5 x 2 x 6²
Kf= 36 J
Initial kinetic energy Ki
Ki = 1/2 m u²
Ki= 0.5 x 2 x 2²
Ki = 4 J
We know that net work is equal to the change in kinetic energy
W(net) = Kf - Ki
W(net) = 36 - 4
W(net) = 32 J
Answer:
friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.
