Can someone please help me ASAP please!

An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th

Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0

3 years ago

1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the

zheka24 [161]

Answer:

a) F = 4.9 10⁴ N, b) F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

∑F = 0

F-W = 0

F = W = mg

F = 5000 9.8

F = 4.9 10⁴ N

the area of the pisto is

A = pi r²

A = pi d² / 4

A = pi 1 ^ 2/4

A = 0.7854 m²

pressure is

P = 4.9 104 / 0.7854

P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

where subscript 1 is for the small piston and subscript 2 is for the large piston

F₁ = $\frac{A_1}{A_2} \ F_2$

the force applied must be equal to the weight of the truck

F₁ = $( \frac{d_1}{d_2} )^2\ m g$

F₁ = (0.05 / 1) ² 5000 9.8

F₁ = 122.5 N

7 0

3 years ago

A net force of 50 newtons is applied to a 20 kilogram cart that is already moving at 1 m/s the final speed of the cart was 3 m/s

valina [46]

Answer:

0.8 seconds

Explanation:

F=ma

Let x be the seconds the force is applied.

m = 20kg

F = 50 Newtons (kg*m/sec^2)

acceleration, a, is provided for x seconds to increase the speed from 1 m/s to 3 m/s, an increase of 2m/s

Let's calculate the acceleration of the cart:

F=ma

(50 kg*m/s^2) = (20kg)*a

a = 2.5 m/s^2

---

The acceleration is 2.5 m/s^2. The cart increases speed by 2.5 m/s every second.

We want the number of seconds it takes to add 2.0 m/sec to the speed:

(2.5 m/s^2)*x = 2.0 m/s

x = (2.0/2.5) sec

x = 0.8 seconds

7 0

3 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 122m in 27s?

Stolb23 [73]

Speed = distance/time
speed= 122÷27=4.52m/s (3sf)

7 0

3 years ago

A projectile is launched from the ground at an angle of 60o above the horizontal. At what point in its trajectory does it have t

Elodia [21]

Answer:

It's constant everywhere in its trajectory.

Explanation:

the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.

The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.

This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.

5 0

3 years ago