Question

A package of mass 56kg is pushed up a ramp with a horizontal force of 450N. The length of the ramp is 3m and the angle between the ramp and the horizontal is 36 degree. If the final speed of the package is 5 m/s, calculate the initial velocity of the package.

Answer 1

Answer:

PLEASE WRITE THE QUESTION CORRECTLY....IT IS VERY DIFFICULT TO UNDERSTAND.

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Answer 2

Given:

Mass of package = 56kg pushed up a ramp with a horizontal force of 450N

Length of ramp =3m  angle between the ramp and horizontal is 36 degree

Final speed of package = 5m/s

To Find:

Initial velocity of the package

Solution:

Free Body Diagram of package on ramp

F = 450N

Fnet = 0

Fcos36 - upwards direction

mgsin36 - downwards direction

FCos36-mgSin36 = ma

450Cos36 - 56(9.81)Sin36=56(a)

364.05 - 323= 56a

a = .73m/$s^{2}$

Use 3rd equation of motion

$v^{2} - u^{2} = 2as$

where s is distance travelled = 3m

v is final velocity and u is initial velocity

substitute the values and solve the equation we get,

u = 4.54m/s

Initial  velocity of the package is 4.54m/s.