✔ 最佳答案
For your first question, component of weight of the object along the inclined plane = mgsin30* = 4(10)sin30* = 20 N
The applied force is in the direction opposite to the component of weight of the object along the plane and the frictional force.
The net force acting on the object = 35 - 10 - 20 = 5 N up the plane
So, there is a net force up the plane, by Newton's 2nd law of motion, there is an acceleration of the object. As the object is initially at rest, it must attain non-zero speed and hence there is a gain of K.E.
Calculation:
Method 1: By Newton's law of motion,
By Newton's 2nd law of motion,
F = ma
5 = 4a
Acceleration, a = 1.25 ms-2
By equation of motion, v2 = u2 + 2as
v2 = 0 + 2(1.25)(20)
Required speed, v = 7.07 ms-1
Method 2: By conservation of energy
By the law of conservation of energy
Work done by the force = work done against friction + Gain in G.P.E. + Gain of K.E.
Fs = fs + mgh + 1/2 mv2
35(20) = (10)(20) + (4)(10)(20sin30*) + 1/2 (4)v2
Required speed, v = 7.07 ms-1
2009-03-31 15:29:05 補充：
There is a net upward force, okay?
Then by the result of Newton's 2nd law of motion, there must be an upward acceleration, that is its upward speed increases. By K.E. = 1/2 mv^2, its K.E. must increase.
2009-03-31 15:29:47 補充：
It is not necessary for an object to lose K.E. by climbing up...
It is because the work done by the force is even larger than the gain in G.P.E. and work done against friction, so the remaining energy goes to gain of K.E.
