This is a bit of a long one so sorry in advance: A 20kg block starts from rest a

Question

This is a bit of a long one so sorry in advance:

A 20kg block starts from rest at top of a 60 degee inclince and slides down the incline. There is a 0.1 coefficient of friction on the ramp. The block then continues to slide along the horizontal surface where is a 0.3 coefficient of friction. The block then slows down due to friction, until it comes to rest at point B.

A) what is the magnitude of the force due to gravity (weight) of the block on the incline?

B) What is the magnitude of the normal force on the block on the incline?

C) What is the magnitude of the frictional force on the block on the incline?

D) What is the magnitude of the acceleration of the block down the incline?

E) What is the velocity of block at Point A?

F) What is the distance between point A and point B?

I keep mixing my equations up which only makes me more confused. Any help would be helpful, thanks!

Explanation / Answer

A) force due to gravity = mg = 20 x 9.81 = 196.2 N

compenent along the incline = mgsin60 = 169.91 N

B) perpendicular to the incline,

N - mgcos@ = 0

N = 20 x 9.81 x cos60 =98.1 N

C) frictional force f = uN = 0.1 x 98.1 = 9.81 N

D) using Fnet = ma

mgsin@ - f = ma

169.91 - 9.81 = 20a

a = 8 m/s^2

E) distance along the incline, L = h / sin60 = 10 / sin60 = 11.55 m

v^2 -u^2 = 2ad

v^2 - 0 = 2 x 8 x 11.55

v =13.59 m/s

F) friction on horizontal rough surface = umg = 0.3 x 20 x 9.81 = 58.86 N

now decelaration = - ug = -0.3 x 9.81 = - 2.94 m/s^2

using v^2 - u^2 = 2ad

0 - 13.59^2 = 2 x -2.94 x d
d = 31.38 m

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