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3 years ago

What is the difference in forces and displacement when comparing positive and negative work?

Physics

1 answer:

8_murik_8 [283]3 years ago

7 0

Answer:

Explanation:

Work is the product of force and displacement. Work is positive if force and displacement moves in the same direction. Work is negative if force and displacement moves in opposite direction. Take for example, a load is to move from point A to point B in the forward direction. if you apply force on the load from point A to B work done on the load is positive. If it happens that the force you apply is to the opposite direction, the work done will be negative.

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SHOW WORK

Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

$T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\$

For Point b:

$E=\frac{1}{2} m a^2 w^2\\\\$

$=\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J$

For Point C:

$V_{max}= a w$

$= (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}$

For point D:

$X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm$

5 0

3 years ago

If a book has a mass of 3 kg, what is the book's weight in N?

sesenic [268]

Answer:

29.4 N

Explanation:

F = ma

F = (3 kg) (9.8 m/s²)

F = 29.4 N

5 0

3 years ago

If an object is projected upward from ground level with an initial velocity of 144144 ft per sec, then its height in feet after

Liula [17]

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

$v_0 = 144 ft/s$

The equation that describes its height at time t is

$s(t) = -16t^2 + 144 t$ (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

$v(t) = s'(t) = -32t +144$ (2)