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Levart [38]
3 years ago
6

In an experiment, a variable, position-dependent force FC) is exerted on a block of mass 1.0 kg that is moving on a horizontal s

urface. The frictional force between the block and the surface has a constant magnitude of Ff. In addition to the final velocity of the block, which of the following information would students need to test the hypothesis that the work done by the net force on the block is equal to the change in kinetic energy of the block as the block moves from 2 = 0 to z = 5 m?
A The function F(2) for 0 < x < 5 and the value of Fr.

B) The function a(t) for the time interval of travel and the value of F.

C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of 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 Ft.

E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F.
Physics
1 answer:
marshall27 [118]3 years ago
8 0

Answer:

C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Fr.

Explanation:

Yo want to prove the following equation:

W_N=\Delta K\\\\

That is, the net force exerted on an object is equal to the change in the kinetic energy of the object.

The previous equation is also equal to:

F(x)x-F_f=\frac{1}{2}m(v_f^2-v_o^2)    (1)

m: mass of the block

vf: final velocity

v_o: initial velocity

Ff: friction force

F(x): Force

x: distance

You know the values of vf, m and x.

In order to prove the equation (1) it is necessary that you have C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F.  Thus you can calculate experimentally both sides of the equation.

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A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude
Nadya [2.5K]

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<h2>a) S = \frac{1}{2}gt^2\\</h2><h2>b) 6secs</h2><h2>c) 192ft</h2>

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

S = ut + \frac{1}{2}at^{2}

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\

a) An expression for the altitude of the ballast after t seconds is therefore

S = \frac{1}{2}gt^2\\

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576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs

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c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft

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