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Montano1993 [528]
1 year ago
15

The 10-lb block has a speed of 4 ft/s when the force of f=(8t2)f=(8t2) lb is applied. determine the velocity of the block when t

== 2 s. the coefficient of kinetic friction at the surface is μk=0. 2?
Physics
1 answer:
KatRina [158]1 year ago
5 0

The velocity of the block when t == 2 s is 60.7 ft./sec.

Equations of Motion.

Here the friction is F_f = \mu_k N = 0.2 N

+ \uparrow \sum F_y = ma_y; \quad N – 10 = \frac { 10 } { 32.2 }(0) \quad N = 10 lb \\ \begin{aligned} \underrightarrow{ + } \sum F_x = ma_x; \quad 8t^2 – 0.2(10 &) = \frac { 10 } { 32.2 }a \\ & a = 3.22(8t^2 – 2) ft/s^2 \end{aligned}

Kinematics.

The velocity of the block as a function of t can be determined by

integrating dv = adt using the initial condition v = 4 ft./s at t = 0.

\int_{ 4 ft/s }^{ v } dv = \int_0^t 3.22(8t^2 – 2)dt \\ \begin{aligned} v – &4 = 3.22 (\frac 8 3 t^3 – 2t) \\ & v = \{8.5867t^3 – 6.44t + 4 \} ft/s \end{aligned}

The displacement as a function of t can be determined by integrating

ds = vdt using

the initial condition s = 0 at t = 0

\int_0^s ds = \int_0^t (8.5867t^3 – 6.44t + 4)dt \\ s = \{2.1467t^4 – 3.22t^2 + 4t \} ft

at t = 2 sec

s = 30 ft.

Thus, at s = 30 ft.,

\begin{aligned} v &= 8.5867(2.0089^3) – 6.44(2.0089) + 4 \\ &= 60.67 ft/s \\ &= 60.7 ft/s \end{aligned}

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.

Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.

Learn more about kinematics here : brainly.com/question/24486060

#SPJ4

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3 years ago
The two blocks in oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip when the ampli
Setler79 [48]

Answer:

0.72

Explanation:

T = Time period of oscillation = 1.5 s

Angular frequency is given as

w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s

A = Amplitude of oscillation = 40 cm = 0.40 m

\mu = Coefficient of static friction = ?

a = acceleration of the block

m = mass of the block

Maximum acceleration of the block is given as

a = Aw^{2}

frictional force is given as

f = \mu mg

As per newton's second law

f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72

8 0
3 years ago
A hiker walks due East for a distance of 6 km
solniwko [45]

Resultant displacement between the base camp and the cave will be 7.211 km

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Her resultant displacement can be calculated using Pythagoras theorem

H^{2} = P^{2} + B^{2}

     = 4^{2} + 6^{2}

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H = \sqrt{52} = 7.211 km

Resultant displacement between the base camp and the cave will be 7.211 km

To learn more about displacement here :

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3 0
1 year ago
Two thin rectangular sheets (0.16 m x 0.44 m) are identical. In the first sheet the axis of rotation lies along the 0.16-m side,
malfutka [58]

Answer:

t'=1.124s

Explanation:

I_b*a_b=I_a*a_a

a_a=\frac{w_f}{t'}

a_b=\frac{w_f}{t}

I_b*\frac{w_f}{t}=I_a*\frac{w_f}{t'}

I_b*\frac{1}{t}=I_a*\frac{1}{t'}

I=\frac{1}{3}*M*r^2

I_a=\frac{1}{3}*M*r_a^2

I_b=\frac{1}{3}*M*r_b^2

\frac{1}{3}*M*r_b^2*\frac{1}{t}=\frac{1}{3}*M*r_a^2*\frac{1}{t'}

t'=(\frac{r_a}{r_b})^2 *t

t'=(\frac{0.16m}{0.44m})^2*t

t'=\frac{16}{121}*8.5s

t'=1.124s

7 0
3 years ago
(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed
stiks02 [169]

a) 7 buses

b) 6.8 m

Explanation:

a)

The motion of the motorcycle is a projectile motion, so it consists of 2 independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion along the vertical direction

The initial components of the velocity of the motorcycle are:

v_x = u cos(32^{\circ})=(40.0)(cos 32^{\circ})=33.9 m/s\\v_y = u sin(32^{\circ})=(40.0)(sin 32^{\circ})=21.2 m/s

The equation for the vertical motion of the motorcycle is

y=h+u_y t - \frac{1}{2}gt^2

where

y is the altitude at time t

h is the initial height

g=9.8 m/s^2 is the acceleration due to gravity

The bus top is at the same height of the initial ramp, so we have

y=h

And therefore, we can solve the equation for t, to find the time of flight:

0=u_y t - \frac{1}{2}gt^2\\t(u_y-\frac{1}{2}gt)=0\\t=\frac{2u_y}{g}=\frac{2(21.2)}{9.8}=4.33 s

Now we find what is the horizontal distance covered by the motorcycle in its jump, which is given by:

d=v_x t = (33.9)(4.33)=146.8 m

And since each bus has a length of L = 20.0 m, the number of buses that the motorcycle can clear with its jump is:

n=\frac{d}{L}=\frac{146.8}{20}=7.34

So, 7 buses.

b)

In the previous problem, we saw that the total range of the motion of the motorcycle is

d=146.8 m

And we said that this corresponds to 7 buses.

Each bus has a length of

L = 20 m

So, the total length of 7 buses is

L' = 7L=7(20)=140 m

Therefore, the range of the motorcycle is greater than the length of the buses by:

\Delta x = d-L'=146.8-140 = 6.8 m

which means he will miss the last bus by 6.8 meters.

6 0
3 years ago
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