Overcoming Friction in of a sliding box

[Fins_Wish]

Okay, i can't figure this out, can anyone help me?
here is the problem:

A 75 kg box slides down a 25.0 degree ramp with an acceleration of 3.60 m/s squared.
a. find mew k between the box and the ramp.
b. what acceleration would a 175 kg box have on this ramp?



so i got this basic info:
m=75 kg
a=3.60 m/s squared
Force of Gravity= 735.75 N (75 x 9.81)
Theta (angle)= 25
Net Force of Gravity= 270 N

but i have no clue what to do with it all.

can someone please help me tonight?

 

[Fellow]

hi, its best to draw out the free body diagram to look at the forces acting on the block.
since the block is on a slope, a portion of its weight acts normally (perpendicularly) to the slope while another portion acts in a direction parallel to the slope.

the 1st portion, the one acting perpendicularly, it will cause a kinetic friction force to act on the block. using trigonometry, this portion of the weight is given by 75*g*cos(25). hence, the kinetic frictional force acting on the block is u_k*75*g*cos(25), where u_k is the coefficient of kinetic friction between the block and the surface.

for the 2nd portion of the weight, the one acting parallel to the surface of the slope, it will cause the block to accelerate down the slope. Again, using trigonometry, this portion of the weight is given by 75*g*sin(25).

since it is given that the block accelerates down the slope at 3.60ms^-2,
the resultant force acting on the block is (3.60)*75 = 75*g*sin(25) - u_k*75*g*cos(25) .. then just solve the equation and get the value of u_k.