The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied
1 answer:
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Answer:
The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).
(C) is correct option.
Explanation:
Given that,
Mass of block = 1.0 kg
Dependent force = F(x)
Frictional force = F(f)
Suppose, the following information would students need to test the hypothesis,
(A) The function F(x) for 0 < x < 5 and the value of F(f).
(B) The function a(t) for the time interval of travel and the value of F(f).
(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).
(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).
(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).
We know that,
The work done by a force is given by,
.....(I)
Where, = net force
We know, the net force is the sum of forces.
So,
According to question,
We have two forces F(x) and F(f)
So, the sum of these forces are
Here, frictional force is negative because F(f) acts against the F(x)
Now put the value in equation (I)
We need to find the value of
Using newton's second law
...(II)
We know that,
Acceleration is rate of change of velocity.
Put the value of a in equation (II)
Now, the work done by the net force on the block is,
The work done by the net force on the block is equal to the change in kinetic energy of the block.
Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).
(C) is correct option.