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Mashutka [201]
3 years ago
13

A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl

ine and has a tension of 140 N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. How fast is the wagon going after moving 80 m up the hill? (THe Answer my teacher gives us is 7.90 m/s, but I have no idea how you get it)
Physics
1 answer:
sp2606 [1]3 years ago
5 0

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

\theta=18.5^{\circ}

Tension=140 N

Initial velocity of wagon=u=0

Displacement=s=80 m

Net force applied  on wagon=F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N

By using g=9.8m/s^2

a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2

We know that

v^2-u^2=2as

Using the formula

v^2=2\times 0.39\times 80

v=\sqrt{2\times 0.39\times 80}=7.9m/s

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Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

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Explanation:

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       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

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            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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