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AlladinOne [14]
3 years ago
13

As a football player moves in a straight line [displacement (3.00 mm)i^i^ - (6.50 mm)j^j^], an opponent exerts a constant force

(126 NN)i^i^ (168 NN)j^j^ on him. How much work does the opponent do on the football player
Physics
2 answers:
LuckyWell [14K]3 years ago
6 0

Answer:

0.714 Joule

Explanation:

Work done is defined as the dot product of force vector and the displacement vector.

W=\overrightarrow{F}.\overrightarrow{d}

Here, \overrightarrow{d} = (3\widehat{i}-6.5\widehat{j})mm=(0.003\widehat{i}-0.0065\widehat{j})m

\overrightarrow{F}=(126\widehat{i}+168\widehat{j})N

W=(126\widehat{i}+168\widehat{j}).(0.003\widehat{i}-0.0065\widehat{j})

W = 0.378  - 1.092

W = - 0.714 Joule

Thus, the work is 0.714 Joule.

Brilliant_brown [7]3 years ago
4 0

Answer:

Work done, W=(0.378i-1.092j)\ J

Explanation:

Displacement,

d=(3i-6.5j)\ mm\\\\d=(0.003i-0.0065j)\ m

Force, F=(126i+168j)\ N

Work done by the opponent do on the football player is given by :

W=F{\cdot} d\\\\W=(126i+168j){\cdot} (0.003i-0.0065j)\ m\\\\W=(0.378i-1.092j)\ J

So, the work done by the opponent do on the football player is  (0.378i-1.092j)\ J.

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Answer:

v_{0}=319.2 m/s    

Explanation:

We need to use the momentum and energy conservation.

p_{0}}=p_{f}

mv_{0}=(m+M)V_{1}

Where:

  • m is the mass of bullet (m=0.01 kg)
  • M is the mass of the wooden (M=0.75 kg)
  • v(0) initial velocity of bullet
  • V(1) is the velocity of the combined object in the moment the bullet hist the block

Conservation of energy.

We have kinetic energy at first and kinetic and potential energy at the end.            

(1/2)(m+M)V_{1}^{2}=(1/2)(m+M)V_{2}^{2}+(m+M)gh

Here:

  • V(1) is the velocity of the combined at the initial position
  • h is the vertical height ( h = 0.800 m)

We can find V(2) using the definition of force at this point:

\Sigma F=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

T-(m+M)gcos(\theta)=(m+M)a_{c}=(m+M)(V_{2}^{2}/R)

cos(\theta) =(L-h)/L=(1.72-0.8)/1.72

\theta =57.66

Now, we can solve the equation to find V(2)

V_{2}=\sqrt{\frac{R*(T-(m+M)*g*cos(\theta))}{(m+M)}}

V_{2}=\sqrt{\frac{1.72*(4.86-(0.01+0.75)*9.81*cos(57.66))}{(0.01+0.75)}}

V_{2}=1.40 m/s        

Now we can find V(1) using the conservation of energy equation

(1/2)V_{1}^{2}=(1/2)V_{2}^{2}+gh

V_{1}=\sqrt{V_{2}^{2}+2gh}

V_{1}=\sqrt{1.40^{2}+2*9.81*0.8}          

V_{1}=4.20 m/s        

Finally, using the momentum equation we find v(0)

v_{0}=\frac{(m+M)V_{1}}{m}                

v_{0}=\frac{(0.01+0.75)*4.20}{0.01}

v_{0}=319.2 m/s        

I hope it helps you!

 

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