The answer is no. If you are dealing with a conservative force and the object begins and ends at the same potential then the work is zero, regardless of the distance travelled. This can be shown using the work-energy theorem which states that the work done by a force is equal to the change in kinetic energy of the object.
W=KEf−KEi
An example of this would be a mass moving on a frictionless curved track under the force of gravity.
The work done by the force of gravity in moving the objects in both case A and B is the same (=0, since the object begins and ends with zero velocity) but the object travels a much greater distance in case B, even though the force is constant in both cases.
Answer:
R = -2500N
Explanation:
Use formula ΣF = Ma. We can split the Net Force into two parts: Fe - Ff = Ma.
Fe is the force developed by the engine: 4000N
Ff (R) is the air resistance projected on the vehicle. This is the value we're looking for.
Mass: 1000kg
Acceleration: 1.5m/s^2
Next, just plug in the values and solve.
▪4000N - Ff = (1000kg)(1.5m/s^2)
▪Ff = 1500N - 4000N
▪Ff = -2500N
R = -2500N.
The air resistance acting on the car is R = -2500N.
Given,
The force applied to lift the tree, F=150 N
The work done in lifting the tree, W=250 J
The work done in lifting the tree is the product of the force applied and the height through which the tree was lifted.
Thus,
Where d is the height through which the tree was lifted.
On substituting the known values,
Thus the tree was lifted through a height of 1.67 m
Answer:
5 m
Explanation:
30 m/s / 6.0 cycles/s = 5 m/cycle
The frequency<span> of oscillation of the dot is the </span>frequency<span> #f# of the </span>wave<span>. The second </span>wave<span> has twice the </span>frequency<span> of the first. </span>You<span> can see that the </span>wavelength<span>is halved when the </span>frequency<span> is doubled. Since #v# is constant, </span>if<span> #f# increases, #λ# must decrease, and vice versa</span>