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Elena L [17]
3 years ago
15

While riding TRAX at a constant speed of 16 m/s you walk toward the front of the car at 5 m/s.

Physics
1 answer:
AVprozaik [17]3 years ago
4 0

Speed of the TRAX is given as 16 m/s so let say it is given as

v_t = 16 m/s

now our speed towards the front end is given by 5 m/s so this is the relative speed of us with respect to TRAX

let say this speed is given as

v_1t = 5 m/s

now we need to find the speed with respect to someone standing outside the TRAX

so here we need to find the net speed in ground frame and hence we can use the formula of relative speed

v_{1t} = v_1 - v_t

v_1 = v_{1t} + v_t

v_1 = 5 + 16

v_1 = 21 m/s

so someone outside the TRAX will see our speed as 21 m/s

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A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j
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3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
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Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

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