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iogann1982 [59]
3 years ago
7

A dog pushes a toy horizontally on a frictionless floor with a net force of 2.0\, \text N2.0N for 1.0\,\text m1.0m. How much kin

etic energy does the toy gain?
Physics
2 answers:
TiliK225 [7]3 years ago
6 0

Answer: 2 J

Explanation: Kinetic energy is possessed by a body due to virtue of its motion. A body gains kinetic energy when some work is done on it.

W = Δ K.E.

We know that work is said to be done on a body when a force causes displacement of the body.

W = F.s

It is given that net force on the toy, F = 2.0 N which displaces the toy by, s = 1.0 m.

Hence, work done on the toy, W = 2.0 N × 1.0 m = 2 J which is equal to the kinetic energy gained by the toy.  

IRINA_888 [86]3 years ago
4 0

Answer:

2.0 J

Explanation:

Since the surface is frictionless, we can solve the problem by using the work-energy theorem, which states that the work done on the toy is equal to the kinetic energy gained by the toy:

W=\Delta E_k

where

W is the work done on the toy

\Delta E_k is the kinetic energy gained by the toy

The work done can be calculated as the product between the force applied on the toy and its displacement, so:

W=Fd=(2.0 N)(1.0 m)=2.0 J

Therefore, according to the work-energy theorem, the toy has gained 2.0 J of kinetic energy.

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A bullet of mass 10g is fired from a gun. The bullet takes 0.003s to move through its barrel and leaves it with a velocity of 30
dsp73

Answer:

1,000 N

Explanation:

v=300 \text{ms}^{-1} ;u=0,t=000.3\text{s},m=0.01\text{kg}

Force exerted by the bullet on the rifle = \frac{m(v-u)}{\text{time}}

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Contextual Way:

Newton's second law of motion.

F=m a

Now to solve based on the current info, we shall assume that:-

The force exerted on the bullet was uniform across the entire duration of bullet leaving the barrel, i.e., 0.003 seconds {Not necessarily true for real life applications as the force will not be uniform from the point of hammer impact till the point of leaving the barrel. In reality you will get a Force profile across that entire duration}

We are not distinguishing between bullet and cartridge. {What you shall hold in your hand and load in a revolver is a cartridge containing the gunpowder, bullet etc. The bullet is the projectile at the mouth of the cartridge that actually leaves the barrel and hit the target. So when you are weighing in real life, you are not weighing the bullet, rather the cartridge as a whole}

Getting back to the question

Impulse equation for the bullet

∫F∗dt=∫m∗dv

Average impulse delivered= Change in momentum of the bullet

Assuming average force delivery

Favg∗∫dt=m∗∫dv

Favg∗0.003=0.010∗300

Favg=300∗10/3

Favg=1000N

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