What happens to the kinetic energy of a snowball as it rolls across the lawn and gains mass.

2 answers:

soldier1979 [14.2K]4 years ago

7 0

The kinetic energy of an object is
$\frac{1}{2} mv^2$
where m is the mass of the object
and v is the velocity.

If you roll a snowball downhill, gravity accelerates the object and v increases. This causes the kinetic energy of the object to increase as well.

If you roll a snowball on level ground or uphill, friction will cause the velocity to decrease. Thus kinetic energy will decrease as well.

LiRa [457]4 years ago

5 0

If the ground is level, then the snowball can never have
any more kinetic energy than it hand when it left your hand.

If more mass sticks to it as it makes its way across the lawn,
then it must slow down, so that its

KE = (1/2) (present mass) (present speed)²

never exceeds the KE you gave it when you tossed it.

And we're not even talking yet about all the energy it loses
by scraping through the snow and mashing down the blades
of grass in its path.

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A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)