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3 years ago

What is the work-energy theorem equation?

Physics

2 answers:

Lisa [10]3 years ago

8 0

The formula for net work is net work = change in kinetic energy = final kinetic energy - initial kinetic energy.

grin007 [14]3 years ago

7 0

Answer:

W = Fd = KE =1/2mv²

Explanation:

not sure if that's what your looking for but i'm pretty sure this is it.

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Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0

3 years ago

A golf club with 65J of kinetic energy strikes a stationary golf ball with a mass of 46g. The energy transfer is only 20% effici

umka21 [38]

Kinetic energy of golf club = 65J,
kinetic energy supplied to golf ball = 20% of 65 = 0.2 * 65 = 13J,
kinetic energy of ball = [mass * Velocity²]/2,
mass = 46gm = 0.046Kg,
[0.046 * V²]/2 = 13, or 0.046 *V² = 26,
V² = 26/0.046 = 565.22,
V = 23.77 m/sec = initial velocity of golf ball after hitting.

4 0

3 years ago

A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is th

BartSMP [9]

<u>Answer:</u>

The height of ramp = 124.694 m

<u>Explanation:</u>

Using second equation of motion,

$s = ut + \frac{1}{2}at^2$