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3 years ago

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A 6.0-kg box slides down an inclined plane that slopes 39° to the horizontal. The box accelerates at a rate of 3.1 m/s^2. What i

s the coefficient of kinetic friction between the box and the surface of the inclined plane?
0.55
0.40
0.16
0.48

1 answer:

Lynna [10]3 years ago

4 0

I'm pretty sure its b. 0.40 no doubt about it XD

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A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

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Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope ( $F_{t}$ ), vertical gravitational force ( $F_{g}$ ) and vertical normal force ( $F_{n}$ ), due to the surface. Since there is no vertical movement, $F_{g}$ and $F_{n}$ cancels it out. So, for this block, net force is horizontal due to the rope $F_{t}$ .

The block of mass m is hanging from the pulley, so there is the force of the rope ( $F_{t}$ ) and the gravitational force ( $F_{g}$ ). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

$F_{r}=m.a$

$a=\frac{F_{r}}{m}$

$a=\frac{18.8}{3.6}$

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

$F_{r}=m.a$

$F_{t}-F_{g}=m.a$

$F_{t}-m.g=m.a$

$18.8-9.8m=5.22m$

$15.02m=18.8$

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: $F_{g}=m.g$ .

The weight for the hanging block is

$F_{g}=1.25*9.8$

$F_{g}=$ 12.25 N

Comparing tension and weight:

$\frac{12.25}{18.8}$ ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

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A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

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Answer:

She will make the jump.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

$77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds$

So she will cover 77 m in 2.85 seconds

Now considering vertical motion, up direction as positive

Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 $m/s^2$ , time = 2.85

Substituting

$s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m$

So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

So she will make the jump.

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After traveling for 6.0 seconds, a runner reaches 10m/s. What is the runner's acceleration? Answer: 1.67 m/s2

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