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Tju [1.3M]
3 years ago
5

A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with t

he ground of 35o. The coefficient of static friction between the box and the board is 0.5 and the coefficient of kinetic friction is 0.3 . A) Will the box slide down the ramp when it is at 35o? Prove it! B) How long will it take the box to slide to the other end of the ramp? (if it slides)
Physics
1 answer:
Vlada [557]3 years ago
3 0

Answer:

Explanation:

When the box is on the ramp , component of its weight along the ramp

= mg sinθ

Friction force acting on it in upward direction

=μ mg cosθ

For sliding

μ mg cosθ < mg sinθ

μ cosθ < sinθ

.5 x cos35 < sin35

.41 < .57

So the box will slide

When sliding starts , kinetic friction acts

Net force in downward direction

mgsinθ - μ mg cosθ

acceleration

= gsinθ - μ g cosθ

= 5.62 - .3 x 9.8 x cos35

= 5.62 - 2.4

= 3.22 m /s²

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What are the 7 clues that a chemical change has occurred
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An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 f
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<h2>a) Initial velocity = 83 ft/s</h2><h2>b) Object's maximum speed = 99.4 ft/s</h2><h2>c) Object's maximum displacement = 153.64 ft</h2><h2>d) Maximum displacement occur at t = 2.59 seconds.</h2><h2>e) The displacement is zero when t = 5.70 seconds</h2><h2>f) Object's maximum height = 153.64 ft</h2>

Explanation:

We have velocity

             v(t)= -32t + 83

Integrating

              s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

              46 = -16 x 0²+83 x 0+C

                 C = 46 feet

So displacement is

              s(t) = -16t²+83t+46

a) Initial velocity is

                 v(0)= -32 x 0 + 83 = 83 ft/s

       Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

             0 = -16t²+83t+46

             t = 5.70 seconds

Substituting in velocity equation

           v(t)= -32 x 5.70 + 83 = -99.4 ft/s

           Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

   That is

                 -32t + 83 = 0

                       t = 2.59 s

Substituting in displacement equation

                s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

   The displacement is zero when t = 5.70 seconds

f) Same as option d

   Object's maximum height = 153.64 ft

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