Please show work and explain, I\'m a little stuck on this. A 4.00-kg block is se

Question

Please show work and explain, I'm a little stuck on this.

A 4.00-kg block is set into motion up an inclined plane with an initial speed of v_i = 7.40 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of theta = 30.0 degree to the horizontal. For this motion, determine the change in the block's kinetic energy. For this motion, determine the change in potential energy of the block-Earth system. Determine the friction force exerted on the block (assumed to be constant). What is the coefficient of kinetic friction?

Explanation / Answer

here,

mass of blcok, m = 4 kg

initial velocity, u = 7.40 m/s

distance travelled, d = 3 m

from Third eqn of motion we have :
v^2 - u^2 = 2 * a * d

final velocity, v = sqrt(u^2 + 2*g*Sin30*d)
final velocity, v = sqrt(7.40^2 + 2*9.81*sin30*3)
final velocity, v = 9.176 m/s^2

part a:
change in ke = initial - final
change in ke = 0.5*m*(v^2 - u^2)
change in ke = 0.5*4*(9.176^2 - 7.4^2)
change in ke = 58.878 J

part b:
from conservation of energy :
change in ke = change in potential energy
change in potential energy = 58.878 J

part c:
From newton second law : SUM(F) = 0
mgSin30 - frictional force = 0

frictional force = mg*Sin30
frictional force = 4*9.81*Sin30
frictional force = 19.62 N

part d:
coefficient of friction = frictional force/normal force

coefficient of friction = Ff/mg*Cos30
coefficient of friction = 19.62/(4*9.81*Cos30)
coefficient of friction = 0.577

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