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Komok [63]
1 year ago
12

two large boxes sit side by side on a sidewalk. the box on the left has a mass of 80kg and the box on the right has a mass of 50

kg. a force is applied to the box on the left , such that both boxes accelerate to the right. draw a free body diagram for each box. include the force of friction
Physics
1 answer:
garri49 [273]1 year ago
8 0

     The force that prevents motion when the surfaces of two objects come into contact is known as friction. Friction decreases a machine's mechanical advantage, or, to put it another way, reduces the output to input ratio.

<h3>How can I figure out the frictional force?</h3>

        The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value.

The formula fr = Fr/N serves as a representation of it.

Therefore, 100N of force is needed to move an item with a mass of 50 kg.

It will accelerate by 10 m/s2.

If a substance's mass does not change over time, friction cannot affect it. Instead, friction can be affected in a variety of ways by an object's mass.

To  Learn more About Friction, Refer:

brainly.com/question/24338873

#SPJ13

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Momentum = mass x velocity so if velocity is kept same and mass is increased then momentum will increase.
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Answer:

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3 years ago
Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?
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Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

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A newton is just \frac{kg·m}{s^{2}}

a=\frac{\frac{10kg·m}{s^{2}}}{50kg}

The s^2 and 50 kg you multiply

a=\frac{10kg·m}{50kg·s^{2}}

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3 years ago
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before
Darina [25.2K]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before  \: it \: comes \: to \: rest \:( s_{2} )}

\\

{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity   = v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \:  s_{1} \: penetration =  \dfrac{v}{2}  \:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{s_{1} =  \dfrac{3}{100}  = 0.03 \: m}

\\

☯ As we know that,

\\

\dashrightarrow\:\: \sf{ {v}^{2}  =  {u}^{2} + 2as }

\\

\dashrightarrow\:\: \sf{  \bigg(\dfrac{v}{2} \bigg)^{2}  =  {v}^{2}   + 2a s_{1}}

\\

\dashrightarrow\:\: \sf{  \dfrac{ {v}^{2} }{4}  =  {v}^{2}  + 2 \times a \times 0.03  }

\\

\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4}  -  {v}^{2}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{\dfrac{ -  3{v}^{2} }{4}  = 0.06 \times a  }

\\

\dashrightarrow\:\: \sf{a =  \dfrac{ - 3 {v}^{2} }{4 \times 0.06}  }

\\

\dashrightarrow\:\: \sf{ a =  \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }

\\

\:\:\:\:\bullet\:\:\:\sf{  Initial\:velocity=v\:m/s} \\\\

\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }

\\

\dashrightarrow\:\: \sf{  {v}^{2}  =  {u}^{2}  + 2as}

\\

\dashrightarrow\:\: \sf{{0}^{2}  =  {v}^{2}  + 2 \times  \dfrac{ - 25 {v}^{2} }{2}  \times s  }

\\

\dashrightarrow\:\: \sf{ -  {v}^{2}  =  - 25 {v}^{2}  \times s  }

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{ -  {v}^{2} }{ - 25 {v}^{2} }}

\\

\dashrightarrow\:\: \sf{  s =  \dfrac{1}{25} }

\\

\dashrightarrow\:\: \sf{ s = 0.04 \: m }

\\

☯ For left penetration (s₂)

\\

\dashrightarrow\:\: \sf{s =  s_{1} +  s_{2}  }

\\

\dashrightarrow\:\: \sf{  0.04 = 0.03 +  s_{2}}

\\

\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }

\\

\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}

\\

\star\:\sf{Left \: penetration \: before  \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\

4 0
3 years ago
Consider a parachutist that has reached terminal velocity. which of the following is true? A.) The acceleration of the parachuti
trasher [3.6K]
The correct answer is:
<span>B.) At terminal velocity there is no net force 

In fact, when the parachutist reaches the terminal velocity, his velocity does not change any more. It means that the acceleration acting on the parachutist is zero, and for Newton's second law, this means the net force acting on him is zero:
</span>\sum F = ma = 0
<span>because the acceleration is zero: a=0. 
This also means that the two relevant forces acting on the parachutist (gravity, downward, and air resistance, upward) are balanced to produce a net force equal to zero.</span>
8 0
3 years ago
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