Mechanics, help?

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1 Answer
May 8, 2018

See below

Explanation:

Let's start out by determining the force of weight, force normal and force parallel down the slope:

F_w= 50kg *9.8m/s^2= 490 N

F_n= 50kg*9.8m/s^2*cos(10^@)=482.6 N

F_p= 50 kg*9.8m/s^2*sin(10^@)=85.1 N

So we have the force parallel that want to pull the box down the ramp, so we know that the force of static friction must be at least equal to force parallel, to make the object not move:

F_n*u_s >= F_p

u_s>=F_p/F_n

u_s>="85.1 N"/"482.6 N"

u_s>=0.176

Ok, so this girl is pushing down the box down the ramp, the force parallel will work in her favor, while the force of friction will not so:

F_F= F_n*u_k

F_F= 482.6 N*0.19= 91.694 N

F_"net"= 50 N+85.1 N-91.694 N= 43.406 N

The work energy conservation theorem states:

W= KE_f-KE_i

The initial KE energy is 0 J as the box is not moving so:

(43.406 N)(5 m)= 1/2(50 kg)v^2

v= 2.95 m/s

The third part I'm not entirely sure of but, 20^@ below the horizontal would still mean that the inclination of the ramp, from the bottom of the ramp at the other horizontal is 20^@

a= F_"net"/m

F_p= 50 kg*9.8m/s^2*sin(20^@)=167.59 N

F_n= 50 kg*9.8m/s^2*cos(20^@)=460.45 N

F_f= 460.45*0.19= 87.486 N

F_"net"= 167.59 N-87.486 N= 80.104 N

a= "80.104 N"/"50 kg"

a= 1.6 m/s^2