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Sloan [31]
3 years ago
9

A box is sliding down an incline tilted at a 16.0 angle above horizontal. The box is initially sliding down the incline at a spe

ed of 1.90 m/s. The coefficient of kinetic friction between the box and the incline is 0.420. How far does the box slide down the incline before coming to rest?
A) 2.33 m

B) 0.720 m

C) 1.78 m

D) 1.44 m

E) The box does not stop. It accelerates down the plane
Physics
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

d) 1.44m

Explanation:

According to Newton's second law:

\sum F=m.a\\

analyzing the horizontal components of the force:

\sum F_x=mg*sin(\theta)-\µ*m.g*cos(\theta)\\\sum F_x=m*9.8*(sin(16^o)-0.420*cos(16^o))\\\sum F_x=-1.25m

applying the second law

-1.25m=m.a\\a=-1.25m/s^2

given the acceleration we can calculate the distances traveled before stopping:

(v_f)^2=(v_o)^2+2.a.\Delta x\\0=(1.90m/s)^2-2(1.25)*\Delta x\\\Delta x=1.44m

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A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t
Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

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