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2 years ago

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A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How lon

g will it take the car to reach the bottom of the ramp if there is no friction?

1 answer:

frosja888 [35]2 years ago

4 0

Answer:

t=0.704s

Explanation:

A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?

from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed

$V^{2} =u^{2} +2as$

Vy^2=0+2*9.8*1.73sin40.5

Vy^2=22.021

Vy=4.69m/s

Vx^2=u^2+2*9.81*cos40.5

Vy^2=25.81

Vy=5.08m/s

V=(Vy^2+Vx^2)^0.5

V=47.71^0.5

V=6.9m/s

from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.

f=force applied

v=velocity final

u=initial velocity

m=mass of the toy, 0.046

f=ma

f=m(v-u)/t

v=u+at

6.9=0+9.8t

t=6.9/9.81

t=0.704s

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For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

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Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

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$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m (or 0.80 m)

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I Hope this helps, greetings ... Dexteright02! =)

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