A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

Answer:

The BTU, or British thermal unit, is actually a measure of heat.

The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.

The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.

From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.

where;

∴

Using trigonometric approach for ΔOAP;

[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]

[tex]\mathbf{AP =OP\times sin \theta}}[/tex]

AP = 200 × sin 30°

Learn more about projectile here:

https://brainly.com/question/20689870?referrer=searchResults

Answer:

1.25 m

Explanation:

This is the vertical height not the distance along the slope.

[tex]K=U\\\frac{1}{2}mv^{2} = mgh\\h = \frac{v^{2}}{2g}=\frac{25}{20}=1.25 m[/tex]

The height the car can reach if the the track starts to climb upwards is 1.2742 meters up.

Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or its relation with its surrounding systems.

P.E. = mass × g × height

K.E. = 0.5 × mass × (velocity)²

Given that the toy car has a mass of 0.025 kg and is traveling on a horizontal track with a velocity of 5 m/s. Now, the car starts to climb up vertically, therefore, the kinetic energy will be converted to potential energy.

Kinetic Energy = Potential Energy

0.5 × mass × (velocity)² = mass × g × height

Cancel mass from both the sides,

0.5 × (velocity)² = g × height

0.5 × (5 m/s)² = 9.81 m/sec² × height

height = 1.2742 meters

Hence, the car will travel 1.2742 meters up.

Learn more about Kinetic and Potential Energy here:

https://brainly.com/question/15764612

#SPJ5

Answer:

a) The ideal speed = 16.21 m/s

b) Minimum co-efficient of friction = 0.216

Explanation:

From the given information:

The ideal speed can be determined by considering the centrifugal force component and the gravity component.

[tex]\dfrac{mv^2}{r}cos \theta = mg sin \theta[/tex]

[tex]v = \sqrt {gr \ tan \theta}[/tex]

[tex]= \sqrt{(9.8 \ m/s^2) (100) \ tan 15^0}[/tex]

= 16.21 m/s

(b)

Let assume that it requires 25 km/h to take the same curve.

Then, using the equilibrium conditions;

[tex]mg \ sin \theta = \dfrac{mv^2}{r} cos \theta + \mu ((\dfrac{mv^2}{r}) sin \theta + mg cos \theta)[/tex]

[tex]\mu = \dfrac{mg sin \theta - \dfrac{mv^2}{r} cos \theta }{((\dfrac{mv^2}{r}) sin \theta + mg cos \theta) }[/tex]

[tex]\mu = \dfrac{g sin \theta - \dfrac{ v^2}{r} cos \theta }{((\dfrac{v^2}{r}) sin \theta + g cos \theta) }[/tex]

[tex]\mu = \dfrac{(9.8 \ m/s^2 ) sin (15^0) - \dfrac{ \dfrac{(25 \times 10^3}{3600} \ m/s)^2 }{100 \ m } cos (15^0) }{((\dfrac{(\dfrac{25 \times 10^3}{3600} )^2}{100}) sin 15^0 + (9.8 \ m/s^2) cos 15^0 ) }[/tex]

[tex]\mathbf{\mu = 0.216}[/tex]