Physics, 15.10.2020 02:01 Kataklusmos

A child tugs on a rope attached to a 0.62-kg toy with a horizontal force of 16.3 N. A puppy pulls the toy in the opposite direction with a force 15.8 N. What is the acceleration of the toy? (Hint: Remember direction: towards child or puppy?)

A child tugs on a rope attached to a 0.62-kg toy with a horizontal force

of 16.3 N. A puppy pulls

Answers: 3

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Answers

Answer from: addietreed2969

In physics, tension is described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar object, or by each end of a rod, truss member, or similar three-dimensional object.

According to the question, the formula we use in this method is F = ma as we have to find the acceleration.

Therefore, the formula for the acceleration is

$a= \frac{F}{m}$

$a= \frac{0.5}{0.62}$

Hence, after the calculation, the acceleration is 0.8m/s.

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Answer from: jeisleen6808

$a=0.8\ m/s^2$

Explanation:

Given that,

Mass of a toy, m = 0.62 kg

Horizontal force, F = 16.3 N

Force on the opposite direction, F' = 15.8 N

We need to find the acceleration of the toy.

Here, two forces are acting in opposite direction, the net force will be the difference of forces.

Net force = 16.3 N-15.8 N

=0.5 N

The formula for net force is given by :

F = ma

a is the acceleration of the toy

$a=\dfrac{F}{m}\\\\a=\dfrac{0.5\ N}{0.62\ kg}\\\\a=0.8\ m/s^2$

So, the acceleration of the toy is $0.8\ m/s^2$ .

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A child tugs on a rope attached to a 0.62-kg toy with a horizontal force of 16.3 N. A puppy pulls t...

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