If a 2.00 kg ball is thrown straight upward with a KE of 500 J.

A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr

irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

time to slide 2 m = ?

a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

$a = \dfrac{T}{m_2} - g$ .......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

$a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}$ .........(2)

from equations 1 and 2

$\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}$

$\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)$

$T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)$

$T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}$

$T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}$

$T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}$

T = 90.61 N

from equation (1)

$a = \dfrac{90.61}{8} - 9.8$ .......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

$t = \sqrt{2.63}$

t = 1.62 s

5 0

3 years ago

Arightward force of 302 N is applied to a 28.6-kg crate to accelerate it across the floor. What will its acceleration be?

Musya8 [376]

Answer:

a $\approx$ 10.6 m/s²

Explanation:

Since F = ma (Force = mass * acceleration), acceleration would be...

a = F/m

a = 302 N/28.6

a $\approx$ 10.6 m/s²

8 0

2 years ago

The position coordinate of a particle which is confined to move along a straight line is given by s =2t3−24t+6, where s is measu

Gre4nikov [31]

Answer:

a) the answer is t=4 seconds

b) acceleration is zero

c) displacement= 142 m

Explanation:

Given the position of the particle

$s=2t^3-24t+6$

a) the time required when velocity $v=72 m/s$

$v=72=\frac{ds}{st}=6t^2-24$

now we solve for time $t$

$6t^2=72+24=96\\\\\implies t^2=\frac{96}{6}\\\therefore t=4 s$

b) acceleration when $v=30 m/s$

acceleration is the time derivative of velocity i.e

$a=\frac{dv}{dt}=\frac{d}{dt}(30)=0$

c) the net displacement of the particle during the interval t = 1 s to t = 4 s is

$s_4-s_1=2t^3-24+6|_{t=4}-2t^3-24+6|_{t=1}\\s_4-s_1=126-(-16)=142 m$

7 0

4 years ago

What is the mass of a bullet moving at 970m/s if the bullet’s KE is 3.9x 10^3J

motikmotik

Answer:

0.08kg

Explanation:

K.E = 1/2 mv^2

v = 970m/s

K.E = 3.9x 10^3J= 3900J

K.E = 1/2 mv^2

3900 = 1/2 m x 970x 970

3900 = 1/2 ×940900m

3900 = 470450m

m = 3900/470450 = 0.00828993516 = 0.008kg

4 0

4 years ago

Read 2 more answers

a rock is vertically upward with a velocity of 10 m/s. calculate the maximum height it reaches and time taken to reach that heig

lina2011 [118]

Answer:

maximum height: p(t) = Vo * t - 1/2 * g * t^2

p’(t) = v(t) = 0 = Vo - g*t. So, maximum height occurs when t = Vo / g

p(Vo / g) = Vo^2/g - 1/2 * g * (Vo/g)^2

Vo = 10 m / s. Let’s approximate g = 10 m / s^2

p(Vo / g) = 10^2 / 10 - 1/2 * 10 * (10/10)^2 = 10 - 5 = 5 meters (approximately)

Calculation of time:

v = u + gt

0 = 10√2 + (-10)t

-10√2 = -10t

2 = √2s

8 0

3 years ago