Question

Mechanics, help?

Answer 1

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`