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4 years ago

A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 m/s. how much work is required to stop it

1 answer:

3 0

One way of thinking of energy is to say the energy is the ability of the body to do work. If you have a body that is moving that body has kinetic energy, as said above this energy can be used to do work. The amount of work done is equal to the amount of energy that body has.
In our case cylinder has the following kinetic energy:
$E_k=m\frac{v^2}{2}\\
E_k=40\frac{6^2}{2}\\
E_k=40\cdot18\\
E_k=720$J$
And that is exaclty how much work you would have to put in to stop the cylinder.

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4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistanc

muminat

a) $15.2 m/s^2$

b) 1.96 N

c) 1.96 N

Explanation:

To find the acceleration of the box, we apply Newton's second law of motion:

$\sum F=ma$

where

$\sum F$ is the net force on the box

m is the mass of the box

a is its acceleration

Here we have to consider the horizontal direction, which is the one in which the box is moving. The net force is given by:

$\sum F=3.55 N - 0.52 N=3.03 N$

which is the difference between the forward force and the resistive force. Then we have

m = 0.200 kg (mass of the box)

Therefore, the acceleration of the box is:

$a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2$

The gravitational force (also called weight) is the force with which an object is pulled by the Earth towards the Earth's centre.

It is given by

$F_g = mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m = 0.200 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the gravitational force on the box is:

$F_g=(0.200)(9.8)=1.96 N$

The normal force is the force with which a surface pushes back on an object.

For an object lying on a flat surface, there are two forces acting along the vertical direction:

- The gravitational force, $F_g$ , which pushes the object downward

- The normal force, N, which pushes the object upward

As the object is at rest, the vertical acceleration of the object is zero, therefore according to Newton's second law of motion, the net force must be zero:

$\sum F=F_g-N=0$