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Marina CMI [18]

2 years ago

14

At the start of his race, an 81-kg runner pushes against the starting block, exerting an average force of 1600 N . The force tha

t the block exerts on his foot points 20∘ above the horizontal. Determine the horizontal speed of the runner after the force is exerted for 0.28 s.

1 answer:

balandron [24]2 years ago

3 0

The force in the horizontal direction is given by 1600cos(20)=1503.5N
Given the mass of m=81kg, we can use F=ma to get acceleration
F/m=a
a=1503.5/81=18.56m/s²
The velocity is given by the integral of the acceleration, V=18.56t m/s
since t=0.28s we get
v=18.56(0.28)=5.197m/s

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A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

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Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

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In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

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$\sum F_{x}=0$

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$-W_{x}+f_{s}=0$

when solving for the static friction we get:

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we can substitute this on our sum of forces equation:

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$N\mu_{s}=mg sin \alpha$

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$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

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$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

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