Ask question

Ask question

All categories

Varvara68 [4.7K]

2 years ago

10

A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

n coefficient is 0.1.
Draw a FBD of all the forces acting on the sled.
What is the weight of the sled?
What force is needed to start the sled moving?
What force is needed to keep the sled moving at a constant velocity?

1 answer:

Diano4ka-milaya [45]2 years ago

8 0

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

You might be interested in

A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ

ra1l [238]

Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

$\Delta U = q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

For a proton,

$q=+e=1.6\cdot 10^{-19}C$

And since $\Delta V=100 V$

The energy gained by the proton is

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

For an alpha particle,

$q=+2e=3.2\cdot 10^{-19}C$

Therefore, the energy gained is

$\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J$

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

$q=+e=1.6\cdot 10^{-19}C$

So the energy gained is the same as the proton:

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

6 0

3 years ago

exis [7]

Answer:

She can swing 1.0 m high.

Explanation:

Hi there!

The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

And the potential energy:

PE = m · g · h

Where:

m = mass of Jane.

v = velocity.

g = acceleration due to gravity (9.8 m/s²).

h = height.

Then:

ME = KE + PE

Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:

ME = KE + PE (PE = 0)

ME = KE

ME = 1/2 · m · (4.5 m/s)²

ME = m · 10.125 m²/s²

When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:

ME = KE + PE (KE = 0)

ME = PE

ME = m · 9.8 m/s² · h

Then, equallizing both expressions of ME and solving for h:

m · 10.125 m²/s² = m · 9.8 m/s² · h

10.125 m²/s² / 9.8 m/s² = h

h = 1.0 m

She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).

6 0

3 years ago

How does the height from which you drop the ball relate to the height that the ball bounces back up?

Stels [109]

The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.

7 0

3 years ago

svlad2 [7]

Answer:

10m

Explanation:

The object distance and image distance is the same from the mirror. so the image is 5m behind the mirror.

5+5=10

5 0

2 years ago

What is black body radiation? Explain in detail.

tangare [24]

An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called black-body radiation

hope it helps

3 0

2 years ago