Ask question

Ask question

All categories

2 years ago

13

A 2.00-kg box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope

, and the box moves upward with a velocity magnitude that varies in time according to v(t)=(2.00m/s2)t+(0.600m/s3)t2. Part A What is the tension in the rope when the velocity of the box is 15.0 m/s ?

1 answer:

Alex2 years ago

4 0

Answer:

T = 27.92 N

Explanation:

For this exercise let's use Newton's second law

T - W = m a

The weight

W = mg

The acceleration can be found by derivatives

a = dv / dt

v = 2 t + 0.6 t²

a = 2 + 0.6 t

We replace

T - mg = m (2 + 0.6t)

T = m (g + 2 + 0.6 t) (1)

Let's look for the time for the speed of 15 m / s

15 = 2 t + 0.6 t²

0.6 t² + 2 t - 15 = 0

We solve the second degree equation

t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6

t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2

We take the positive time

t = 3.6 s

Let's calculate from equation 1

T = 2.00 (9.8 + 2 + 0. 6 3.6)

T = 27.92 N

You might be interested in

A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to

daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Explanation:

It is given that,

Mass of cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=4\ kg$

Initial speed of cart 1, $u_1=3\ m/s$

Initial speed of cart 2, $u_2=0$ (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{2\times 3}{(2+4)}$

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

$J=F\times t=m(V-u)$

$J=m(V-u)$

$J=2\times (1-3)$

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0

2 years ago

Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g$

The acceleration is 1.10261 times g

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s$

In mph

$186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph$

The speed of the dragster is 416.17506 mph

5 0

2 years ago

In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t

gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0

2 years ago

A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

Maru [420]

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=m $v_{2}$ +m $v_{1}$

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

4 0

2 years ago

(a). The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 631m

lions [1.4K]

Explanation:

Given that,

Length of the building, l = 631 m

Breadth of the building, b = 707 yards

Height of the building, h = 110 ft

1 meter = 3.28084 feet

631 m = 2070.21 feet

1 yard= 3 feet

707 yards = 2121 feet

(a) The volume of any cuboidal shape is given by :

$V=l\times b\times h$

$V=2070.21\ ft\times 2121\ ft\times 110\ ft$

$V=4.83\times 10^8\ ft^3$

(b) $V=4.83\times 10^8\ ft^3$

$V=(4.83\times 10^8\ ft^3)(\dfrac{1\ m}{3.281\ ft})^3$

$V=1.37\times 10^7\ m^3$

Hence, this is the required solution.

3 0

2 years ago