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2 years ago

A skateboarder starts at 2.75 m/s and accelerates for 8.90 s over a distance of 16.7

Physics

1 answer:

Andrei [34K]2 years ago

3 0

Answer:

1.00 m/s

Explanation:

Given:

Δx = 16.7 m

v₀ = 2.75 m/s

t = 8.90 s

Find: v

Δx = ½ (v + v₀) t

16.7 m = ½ (v + 2.75 m/s) (8.90 s)

v = 1.00 m/s

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otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

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m = $\frac{83}{9.81}$

= 8.46

Now, we will balance the forces along the y-component as follows.

N = W + $F_{2}$

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Now, balancing the forces along the x component as follows.

$F_{1} - F_{r}$ = ma

$20 - F_{r} = 8.46 \times (-0.9)$

$F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

$F_{r} = \mu \times N$

$\mu = \frac{F_{r}}{N}$

= $\frac{7.614}{108}$

= 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

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