Question

the initial kinetic energy of an object moving on a horizontal surface is K. Friction between the object and the surface causes the velocity of the object to decrease uniformly to zero in time t. What is the kinetic energy of the object at time = t/2?​

Answer 1

Answer:

........................

Answer 2

Answer:

Kinetic energy is the energy an object has because of its motion.

If we want to accelerate an object, then we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed. The energy transferred is known as kinetic energy, and it depends on the mass and speed achieved.

Kinetic energy can be transferred between objects and transformed into other kinds of energy. For example, a flying squirrel might collide with a stationary chipmunk. Following the collision, some of the initial kinetic energy of the squirrel might have been transferred into the chipmunk or transformed to some other form of energy.

How can we calculate kinetic energy?

To calculate kinetic energy, we follow the reasoning outlined above and begin by finding the work done, WWW, by a force, FFF, in a simple example. Consider a box of mass mmm being pushed through a distance ddd along a surface by a force parallel to that surface. As we learned earlier

\begin{aligned} W &= F \cdot d \\ &= m · a · d\end{aligned}

W

=F⋅d

=m⋅a⋅d

Huh? I'm lost already.

If we recall our kinematic equations of motion, we know that we can substitute the acceleration if we know the initial and final velocity—v_\mathrm{i}v

i

v, start subscript, i, end subscript and v_\mathrm{f}v

f

v, start subscript, f, end subscript—as well as the distance. What kinematic formula is this?

\begin{aligned} W &= m\cdot d\cdot \frac{v_\mathrm{f}^2-v_\mathrm{i}^2}{2d} \\ &= m\cdot \frac{v_\mathrm{f}^2-v_\mathrm{i}^2}{2} \\ &= \frac{1}{2}\cdot m \cdot v_\mathrm{f}^2 - \frac{1}{2}\cdot m \cdot v_\mathrm{i}^2 \end{aligned}

W

=m⋅d⋅

2d

v

f

2

−v

i

2

=m⋅

2

v

f

2

−v

i

2

=

2

1

⋅m⋅v

f

2

−

2

1

⋅m⋅v

i

2

So, when a net amount of work is done on an object, the quantity \dfrac{1}{2}mv^2

2

1

mv

2

start fraction, 1, divided by, 2, end fraction, m, v, squared—which we call kinetic energy KKK—changes.

\text{Kinetic Energy: } K=\frac{1}{2}\cdot m\cdot v^2Kinetic Energy: K=

2

1

⋅m⋅v

2

start text, K, i, n, e, t, i, c, space, E, n, e, r, g, y, colon, space, end text, K, equals, start fraction, 1, divided by, 2, end fraction, dot, m, dot, v, squared

Alternatively, one can say that the change in kinetic energy is equal to the net work done on an object or system.

W_{net}=\Delta KW

net

=ΔKW, start subscript, n, e, t, end subscript, equals, delta, K

This result is known as the work-energy theorem and applies quite generally, even with forces that vary in direction and magnitude. It is important in the study of conservation of energy and conservative forces.