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4 years ago

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

1 answer:

6 0

<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>

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A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

$E_{k1}+W=E_{k2}\\$

where:

W = work [J]

Ek1 = kinetic energy at initial state [J]

Ek2 = kinetic energy at the final state = 0.

We must remember that kinetic energy can be calculated by means of the following expression.

$\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]$

We know that work is defined as the product of force by distance.

$W=F*d$

where:

F = force [N]

d = distance [m]

But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.

$f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]$

Now solving the equation for the work.

$d=W/F\\d = 50/19.62\\d = 2.54[m]$

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Answer:

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Explanation:

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Question 23

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If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,

с Stay the same

Explanation:

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Mass increases of a body when an object has higher velocity or the speed.
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