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3 years ago

A car and its passengers have a mass of 1200kg it is travelling at 12m/s.

Physics

1 answer:

Natali [406]3 years ago

4 0

Answer:

The increase of kinetic energy is 108,000 J

Explanation:

Kinetic Energy

Is the energy an object has due to its state of motion. It's proportional to the square of the speed and the mass.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

A car has a total mass of m=1,200 kg and travels at v1=12 m/s. Then it increases its speed at v2=18 m/s.

It's required to compute the increase of kinetic energy. We'll calculate both energies K1 and K2 and then subtract them.

$\displaystyle K_1=\frac{1}{2}1,200*12^2=86,400\ J$

$\displaystyle K_2=\frac{1}{2}1,200*18^2=194,400\ J$

The increase of kinetic energy is:

$\Delta K=K_2-K_1 =194,400\ J-86,400\ J$

$\Delta K=108,000\ J$

The increase of kinetic energy is 108,000 J

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photoshop1234 [79]

Since static friction is the minimum force required to just start the motion of a stationary object.

Here if we need to start an object from rest then we required F = 700 N

So for the first part of the above problem Force will be F = 700 N

Now if the box is already moving then we will have to use kinetic friction force between box and floor

now we can write the equation of net force as

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here

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now we will have

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3 years ago

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4 years ago

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

Hoochie [10]

Answer:

e. the air mattress exerts the same impulse, but a smaller net avg force, on the high-jumper than hard-ground.

Explanation:

This is according to the Newton's second law and energy conservation that the force exerted by the hard-ground is more than the force exerted by the mattress.

The hard ground stops the moving mass by its sudden reaction in the opposite direction of impact force whereas the mattress takes a longer time to stop the motion of same mass in a longer time leading to lesser average reaction force.

Mathematical expression for the Newton's second law of motion is given as:

$F=\frac{dp}{dt}$ ............................................(1)

where:

dp = change in momentum

dt = time taken to change the momentum

We know, momentum:

$p=m.v$

Now, equation (1) becomes:

$F=\frac{d(m.v)}{dt}$

∵mass is constant at speeds v << c (speed of light)

$\therefore F=m.\frac{dv}{dt}$

and, $\frac{dv}{dt} =a$

where: a = acceleration

$\Rightarrow F=m.a$

also

$F\propto \frac{1}{dt}$

so, more the time, lesser the force.

& Impulse:

$I=F.dt$

$I=m.a.dt$

$I=m.\frac{dv}{dt}.dt$

$I=m.dv=dp$

∵Initial velocity and final velocity(=0), of a certain mass is same irrespective of the stopping method.

So, the impulse in both the cases will be same.

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